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FORMULA BANK
Horizontal Motion Formulas: x = vxot + ½at^2 vxf = vxo + at x = ½(vxo + vxf)t vxf^2 = vxo^2 + 2 ax Vertical Motion Formulas: y = yo + vyot ½gt^2 vyf = vyo gt y = ½(vyo + vyf)t vyf^2 = vyo^2 2 gy Pythagorean Theorem: a^2 + b^2 = c^2 S.O.H.C.A.H.T.O.A.: 𝐭𝐚𝐧^ ^ =^ 𝒐𝒑𝒑. 𝒂𝒅𝒋. 𝐬𝐢𝐧^ ^ =^ 𝒐𝒑𝒑. 𝒉𝒚𝒑. 𝐜𝐨𝐬^ ^ =^ 𝒂𝒅𝒋. 𝒉𝒚𝒑. To find the angle : ^ =^ 𝐭𝐚𝐧𝟏^ 𝒐𝒑𝒑. 𝒂𝒅𝒋. ^ =^ 𝐬𝐢𝐧 𝟏 𝒐𝒑𝒑. 𝒉𝒚𝒑. ^ =^ 𝐜𝐨𝐬 𝟏 𝒂𝒅𝒋. 𝒉𝒚𝒑.
- A freshman runs off the bleachers and onto the football field as shown in the figure at right. The freshman’s horizontal velocity is 8 m/s. Assume gravity is 10 m/s per second.
- (a.) ___ y = yo ½gt^2 ___ What is the formula for height you would use for this problem (already “adjusted” for the problem situation)? Ans. The formulas for vertical motion that have time in them are y = yo ±vyot ½gt^2 and vyf = ±vyo gt. The first one is for height and the second one for final velocity. We will use the formula for height and modify it for our situation. The freshman has no initial vertical velocity (he has horizontal velocity but not vertical velocity). The freshman is also jumping from an elevated position (10 m) so she has initial height. So your final formula is y = yo ½gt^2 (b.) t = 1.4 1 sec_ In how many seconds would the freshman land on the ground? Ans. G.U.E.S.S. Method. Givens g = 10 m/s^2 Uknown t (time to hit the ground) Equation From Probl. (a.) y = yo ½gt^2 Solve Let’s solve for our variable in question, t : y = yo ½gt^2 yo yo y yo = ½gt^2 2 (y yo) = ½gt^2 2 2 (y yo) = gt^2 g g √ 2 (𝑦−𝑦𝑜) 𝑔 =^ √𝒕 𝟐 √ 2 (𝑦−𝑦𝑜) 𝑔 =^ t Substitute Plugging in our given information into our formula: √ 2 (𝑦−𝑦𝑜) 𝑔 =^ t √ 2 ( 0 𝑚− 10 𝑚) 10 𝑚⁄^ 𝑠 (^2) = t √^2.^04 = t 𝟏. 𝟒𝟏 𝒎/𝒔 = t
PROJECTILE Practice Worksheet Ans. Key
b = (^) opp. a = adj. c = hypot. 10 m
(c.) ____11. 31 m_____ How far (in m) would the freshman have moved horizontally in the time you found in Probl. (b). Ans. G.U.E.S.S. Method. Givens vxo = 8 m/s (given initial horizontal velocity), t = 1.41 s (from Probl. (b.)) Uknown x (horizontal distance travelled by freshman) Equation the formula for horizontal distance: x = vxot + ½at^2 Adjusting for our situation (no horizontal acceleration): x = vxot Solve The formula is already solved for x: x = vxot Substitute Plugging in our given information into our formula: x = vxot = (8 m/s)(1.41 s) = 11. 31 m (d.) _____ vyf = gt ____ What is the formula for vertical velocity (in m/s) you would use for this problem (already “adjusted” for the problem situation)? Ans. The “cousin” formula for vertical velocity is vyf = gt, because the freshman has no initial vertical velocity. (e.) ____14. 14 m/s_____ What would be the freshman’s final vertical velocity (in m/s) right before he lands on the ground? Ans. G.U.E.S.S. Method. Givens g = 10 m/s^2 , t = 1.4 1 sec (from Probl. (b.)) Uknown vyf (freshman’s final vertical velocity) Equation vyf = gt Solve The formula is already solved for v: vyf = gt Substitute Plugging in our given information into our formula: vyf = gt = 10 m/s^2 1.41 sec = 14. 14 m/s (f.) Draw and label the freshman’s velocity vectors at the points indicated. Make sure you draw them to scale. The first one has been done for you. You should draw both vx, the horizontal components, vy, the vertical components, and vr , the resultant. You will be drawing a total of six arrows that are missing below. Also, label the angles that the resultant vectors make with the horizontal. STEP 3: Let’s draw the vertical velocity vectors now according to our scale.
- At t = 0 sec, There is no initial vertical velocity. vyo = 0 m/s
- At t = 1 sec, the vertical velocity is vy = gt = 10 m/s^2 (1 sec) = 10 m/s Using our scale of 0.3375 cm per 1 m/s, 10 m/s 0.3375 cm per m/s = 3.375 cm
- At t = 1.41 sec, the vertical velocity, vy, is 14.14 m/s (ans. to Probl. (e.)). Using our scale of 0.3375 cm per 1 m/s, 14.14 m/s 0.3375 cm per m/s = 4.7725 cm
10 m
t = 0 sec 1 sec 1.41 sec
vxo = 8 m/s
vx = 8 m/s
vx = 8 m/s
Ans. STEP 1: The horizontal velocity does not change throughout so we can just copy the initial horizontal velocity vector and draw it exactly the same way at the other two points STEP 2 : Let’s measure the length of the horizontal velocity vector and set up a proportion to find our scale. The horizontal velocity is given to be 8 m/s and is 2.7 cm long. Set up a proportion: 𝟐. 𝟕 cm 𝟖 𝑚⁄^ 𝑠^
= 𝟎. 𝟑𝟑𝟕𝟓 𝐜𝐦 𝐩𝐞𝐫 𝒎⁄^ 𝒔^.
- At the SMHS vs. Judson football game, the cheerleading squad launches a Rattler t-shirt with the t-shirt shooter into the bleachers from the football field with an initial velocity of 50 m/s at an angle of 40. Assume gravity is 10 m/s^2. (a.) _38.30 m/s **= vxo** ___ What is the initial horizontal velocity (vxo) of the t-shirt (in m/s)? Ans. The initial horizontal velocity vxo can be obtained from taking apart vr, the initial velocity (resultant), into its horizontal and vertical components. G.U.E.S.S. Method. Givens vr = 50 m/s (the diagonal initial velocity), = 40 (the launch angle) Uknown vxo (the horizontal component of the velocity) Equation 𝐜𝐨𝐬 = 𝒂𝒅𝒋. 𝒉𝒚𝒑. Solve 𝐜𝐨𝐬 = 𝒂𝒅𝒋. 𝒉𝒚𝒑. 𝐜𝐨𝐬 = 𝒗𝒙𝒐 𝒗𝒓. 𝒗𝒓 𝐜𝐨𝐬 = 𝒗𝒙𝒐 𝒗𝒓.^ ^ 𝒗𝒓 𝒗𝒓 𝐜𝐨𝐬 = 𝒗𝒙𝒐 16 0 m
Vr = 50 m/s STEP 6 : Use S.O.H.C.A.H.T.O.A, to find the direction (angles) of the resultant velocities: Ans. G.U.E.S.S. Method. Givens at t = 1 sec, vx = 8 m/s, vy = 10 m/s At t = 1.41 sec, vx = 8 m/s, vy = 14.14 m/s Uknown (the launch angle) Equation = 𝐭𝐚𝐧𝟏^ 𝒐𝒑𝒑. 𝒂𝒅𝒋. Solve The formula is already solved for . Substitute Plugging in our given information into our formula: = 𝐭𝐚𝐧𝟏^ 𝒐𝒑𝒑. 𝒂𝒅𝒋. =^ 𝐭𝐚𝐧 𝟏 𝟏𝟎^ 𝒎/𝒔. 𝟖 𝒎/𝒔. = 51.34 = 𝐭𝐚𝐧𝟏^ 𝒐𝒑𝒑. 𝒂𝒅𝒋. =^ 𝐭𝐚𝐧 𝟏 𝟏𝟒.𝟏𝟒^ 𝒎/𝒔. 𝟖 𝒎/𝒔. = 60. 50
10 m
t = 0 sec 1 sec 1.41 sec
vxo= 8 m/s
vx = 8 m/s
vx = 8 m/s
vy = 10 m/s
vr = 12.81 m/s
vy = 14.14 m/s
vr = 16.25 m/s
Substitute 𝒗𝒓 𝐜𝐨𝐬 = 𝒗𝒙𝒐 (𝟓𝟎 𝒎 𝒔 )^ ^ 𝐜𝐨𝐬^ (𝟒𝟎°)^ =^ 𝒗𝒙𝒐 38.30 m/s = vxo
(b.) 32.14 m/s **= vyo** __ What is the initial vertical velocity (vyo) of the t-shirt (in m/s)? Ans. The initial vertical velocity vyo can be obtained from the Pythagorean Theorem or by taking apart vr, the initial velocity (resultant), into its horizontal and vertical components. G.U.E.S.S. Method. Givens vr = 50 m/s (the diagonal initial velocity), = 40 (the launch angle) Uknown vyo (the vertical component of the velocity) Equation 𝐬𝐢𝐧 = 𝒐𝒑𝒑. 𝒉𝒚𝒑. Solve 𝐬𝐢𝐧 = 𝒐𝒑𝒑. 𝒉𝒚𝒑. 𝐬𝐢𝐧 = 𝒗𝒚𝒐 𝒗𝒓. 𝒗𝒓 𝐬𝐢𝐧 = 𝒗𝒚𝒐 𝒗𝒓. ^ 𝒗𝒓 𝒗𝒓 𝐬𝐢𝐧 = 𝒗𝒚𝒐 (c.) 3.214 s = t In how many seconds will the t-shirt reach its maximum height? Ans. By mental math, we can estimate the time right away. If gravity, takes 10 m/s every second off the initial vertical velocity, v yo, which is 32.14 m/s, then it will take a 3 .214 seconds for the vertical velocity to be zero. The vertical velocity is zero at the maximum height. G.U.E.S.S. Method (if you are not good at mental math). Givens vyo = 32.14 m/s (initial vertical velocity from Probl. (b.)), g = 10 m/s^2 Uknown t (time to reach maximum height) Equation v = ±vyo gt. Adjusted to our situation: v = vyo gt (because there is positive initial velocity) Solve We want to find t , the time. Solving: v = vyo gt v = vyo gt vyo vyo v vyo = gt g g v 𝐯𝐲𝐨 −𝐠 =^ 𝐭 (d.) y = 51.65 m What is the maximum height (in m) the t-shirt will go?
The mental math shortcut formula for this was ymax =
𝒗𝒚𝒐(𝒕𝒕𝒐𝒑) 𝟐
𝟑𝟐.𝟏𝟒𝒎 𝒔 (𝟑.𝟐𝟏𝟒 𝒔) 𝟐
Ans. G.U.E.S.S. Method (if you don’t like the shortcut) Givens vyo = 32.14 m/s (initial vertical velocity from Probl. (b.)), g = 10 m/s^2 , t = 3.214 s (time to get to max. height from Probl. (c.)) Uknown y (maximum height) Equation y = yo ±vyot ½gt^2. Adjusted to our situation: y = vyot ½gt^2 (because there is positive initial velocity and no initial height) Solve We want to find y , the height, so we don’t have to move anything around. y = vyot ½gt^2 Substitute y = 32.14 m/s(3.214 s) – ½(10 m/s^2 )(3.214 s)^2 y = 51.65 m (e.) __123.10 m_____ How far (in m) horizontally will the t-shirt have gone when it is at maximum height? Ans. G.U.E.S.S. Method. Givens vxo = 3 8. 30 m/s (horizontal velocity from Probl. (a.)), t = 3.214 s (time at maximum height from Probl. (c.)) Uknown x (horizontal distance) Equation x = vxot + ½at^2. Adjusted to our situation: x = vxot (because there is positive initial velocity and no acceleration) Solve We want to find x , the distance, so we don’t have to move anything around. x = vxot Substitute x = vxot = 38. 30 m/s (3.214 s) = 123. 10 m Substitute 𝒗𝒓 𝐬𝐢𝐧 = 𝒗𝒚𝒐 (𝟓𝟎 𝒎 𝒔 ) 𝐬𝐢𝐧 (𝟒𝟎°) = 𝒗𝒚𝒐
- 14 m/s = vyo Substitute v 𝐯𝐲𝐨 −𝐠
𝟎m s (𝟑𝟐.𝟏𝟒𝐦 𝐬 ) −𝟏𝟎 (^) 𝐬𝐦𝟐^ =^ 𝐭 𝟑. 𝟐𝟏𝟒 𝐬 = 𝐭
