7a. A ball is dropped from the roof of a 23 in tall building. What is the veloci

Question

7a. A ball is dropped from the roof of a 23 in tall building. What is the velocity of the object when it touches the ground? Velocity= 7b. Suppose that the ball is a perfect golf ball, and it bounces such that the velocity as it leaves the ground hats the same magnitude but the opposite direction as the velocity with which it reached the ground. How high will the ball bounce? height reached = 7c. Now suppose instead that the ball bounces back to a height of 19 m. What was the velocity with which it left the ground? velocity =

Explanation / Answer

7A)

initial velocity =u= is zero,

we need to find the final velocity v

acceleration due to gravity = g=9.81m/s2 ( same direction as motion of ball)

given height h=23 m

now, using kinematics equation we have,

V2 -U2 = 2gh

=> v= sqrt (2gh)

=sqrt(2*9.81*23)

=21.242 m/s

7 B)

now, if bounces back with same velocity it should reach the same height,

lets check

now , initial velocity =u =21.242 m/s

g=-9.81 m/s2 (opposite to direction of motion)

final velocity at topmost point =0

so,

V2 -U2 = 2gh

-21.242 2 =-2 *g*h

=>451.26 =2*9.81*h

=>h=23 m

7C)

now given height = 19m

final velocity v=0

we need to find initial velocity =u

V2 -U2 = 2gh

u=sqrt(2gh)

=-19.307 m/s (negative sign for upward motion)

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