Answer:
A.25.096m
B.10m/s
V=22.189m/s
Explanation:
Part A The ball's maximum height above the ground. Part B The ball's speed as it passes the window on its way down. Part C The speed of impact on the ground.
newton's equation of motion
V^2= u^2+2as
a=-g since it is going against gravity
S=h s= distance travelled
h=height
V=final velocity
U=initial velocity
V=o
u=10m/s
0=100+2(-9.81)h
h=100/19.62
h=5.096m
The maximum height above ground level=
Height from thr window to the maximum the ball reached before coming downward+ height from the window to the ground level
H=h+h1
H=5.096+20
25.096m
2.part b
The final velocity as it reach its maximum height upward becomes zero and then the final velocity upward= the initial velocity when coming down
V=U =0
v^2=u^2+2ah
V^2=0+2(9.81)(5.096)
V=9.99917596605
V=10m/s
3. Speed of its impact on the ground
v^2=u^2+2ah
V^2=0+2(9.81)(25.096)
V^2=492.38352
V=22.189m/s
Using the conversion of energy to resolve the questions
Given data :
initial speed of ball = 10 m/s
Distance of ball from ground ( h ) = 20 m
A) Determine the ball's maximum height above ground
we will apply the Newton's motion equation
V² = u² + 2as ------ ( 1 )
where ; u = 0 m/s, v = 0 m/s , a = -9.81 m/s² , s = ?
Insert values into equation ( 1 )
0 = 100 + 2(-9.81 ) * s
s = 5.1 m
∴ Ball's maximum height ( H ) = s + h = 5.1 + 20 = 25.1 m
B) Determine the Ball's speed as it passes the window on its way down
V² = u² + 2as
where ; V = ? , u = 0 , a = 9.81 m/s² , s = 5.1 m
Hence ; V² = 0 + 2 ( 9.81 ) * 5.1 m
∴ V = 10 m/s
C) Determine the speed of impact on the ground
V² = u² + 2aH
where ; V = ? , u = 0, a = 9.81 m/s², H = 25.1 m
insert values into equation above
V² = 0 + 2( 9.81 ) * 25.1
V = √(0 + 2( 9.81 ) * 25.1 )
= 22.19 m/s
Hence we can conclude that The ball's maximum height above ground = 25.10 m, The ball's speed on its way down = 10 m/s , The impact speed on the ground = 22.19 m/s.
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