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Particle Motion in Two Dimensions Model Worksheet 4: Projectile Motion Problems, Exercises of Physics

Practice problems on Projectile Motion with solutions

Typology: Exercises

2020/2021
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1 U6 2D Motion - ws 4 v3.1
Name
Date Pd
Particle Motion in Two Dimensions Model Worksheet 4:
Projectile Motion Problems
1. A metal sphere is launched with an initial velocity of 1.5 m/s as it leaves the ramp. The end
of the ramp is 1.20 m above the floor. Calculate the range of the sphere. (Range is the
horizontal displacement of the projectile.)
cos20 (1.5 ) 1.4
sin20 (1.5 ) 0.51
mm
xss
mm
yi ss
v
v


2
2
22
11
22
2(use solver or quadratic)
-1.2m=(0.51 )t ( 10 )
( 5 ) (0.51 )t 1.2m 0 0.54
1.4 (0.54 ) 0.76
mm
yi ss
mm
s
s
m
xs
y v t a t t
t t s
x v t x s m
2. Now the ramp is tilted downwards and the sphere leaves the ramp at 1.5 m/s as shown below.
The bottom of the ramp is 0.90 m above the floor. Calculate the range of the sphere.
cos15 (1.5 ) 1.45
sin20 (1.5 ) 0.51
mm
xss
mm
yi ss
v
v

2
2
22
11
22
2(use solver or quadratic)
-0.90m=(-0.51 )t ( 10 )
( 5 ) (-0.51 )t 0.90m 0 0.37
1.45 (0.37 ) 0.54
mm
yi ss
mm
s
s
m
xs
y v t a t t
t t s
x v t x s m
1.2 m
y = -1.2 m
vyi = 0.51
m
s
a = g = -10
2
m
s
vx = 1.4
m
s
t = ?
x = ?
vx
vyi
v = 1.5
m
s
2
y = -0.90 m
vyi = -0.51
m
s
a = g = -10
2
m
s
vx = 1.45
m
s
t = ?
x = ?
0.90 m
y
x
vx
vyi
v = 1.5
m
s
15°
pf3
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Download Particle Motion in Two Dimensions Model Worksheet 4: Projectile Motion Problems and more Exercises Physics in PDF only on Docsity!

Name Date Pd

Particle Motion in Two Dimensions Model Worksheet 4:

Projectile Motion Problems

  1. A metal sphere is launched with an initial velocity of 1.5 m/s as it leaves the ramp. The end of the ramp is 1.20 m above the floor. Calculate the range of the sphere. ( Range is the horizontal displacement of the projectile.)

cos 20 (1.5 ) 1. sin20 (1.5 ) 0.

x ms ms yi ms ms

v v

2 2

12 2 12 2 (^2) (use solver or quadratic)

-1.2m=(0.51 ) t ( 10 ) ( 5 ) (0.51 ) t 1.2m 0 0. 1.4 (0.54 ) 0.

yi ms ms sm^ ms x ms

y v t a t t t t s x v t x s m

  1. Now the ramp is tilted downwards and the sphere leaves the ramp at 1.5 m/s as shown below. The bottom of the ramp is 0.90 m above the floor. Calculate the range of the sphere.

cos15 (1.5 ) 1. sin20 (1.5 ) 0.

x ms ms yi ms ms

v v

2 2

12 2 12 2 (^2) (use solver or quadratic)

-0.90m=(-0.51 ) t ( 10 ) ( 5 ) (-0.51 ) t 0.90m 0 0. 1.45 (0.37 ) 0.

yi ms (^) sm sm^ ms x ms

y v t a t t t t s x v t x s m

1.2 m

y = - 1.2 m vyi = 0.51 ms a = g = - 10^ ms 2 vx = 1.4 ms  t =?x =?

y

x

vx

vyi

v = 1.5 ms 2 0°

y = - 0.90 m vyi = - 0.51 ms a = g = - 10^ ms 2 vx = 1.45 ms  t =?x =?

0.90 m  y

x

vx vyi v = 1.5 ms

15 °

  1. A water balloon is launched at a building 24 m away with an initial velocity of 18 m/s at an angle of 50˚ above the horizontal. a. At what height will the balloon strike the building? cos50 (18 ) 11. sin50 (18 ) 13.

x ms ms yi ms ms

v v

12 2 12 2 2

=(13.8 )2.1 ( 10 )(2.1s) 6.

x (^) x ms yi ms ms

x v t t x^ m s v y v t a t  y s m

b. If the balloon misses or shoots over the building, how far will the balloon land from its launch location?

2 2

12 2 0 (13.8 ) 12 ( 10 )^2

yi ms ms sm^ ms x ms

y v t a t t t t t s x v t s m

c. The balloon can be launched from less than 24 m away from the building at the same speed and angle and still hit exactly the same height you calculated in part a. Determine this second launch location.

2 2

12 2 12 2 (^2) (use solver or quadratic)

6.9m=(13.8 ) t ( 10 ) ( 5 ) (13.8 ) t 6.9m 0. 11.6 (0.66 ) 7.

yi ms ms sm^ ms x ms

y v t a t t t t s x v t x s m

building vx

v vyi 50°yx vyi = 13.8 ms a = g = - (^10) s^ m 2 vx = 11.6 ms  x = 24 mt =?y =?

vyi = 13.8 ms a = g = - 10^ ms 2 vx = 11.6 ms  y = 0t =?x =?

vyi = 13.8 ms a = g = - (^10) s^ m 2 vx = 11.6 ms  y = 6.9 mt =?x =?