A tennis ball is thrown vertically upward with an initial velocity of +7.7 m/s. What will the ball's velocity be when it returns to starting point?Answer in units of m/s. How long will the ball...

A tennis ball is thrown vertically upward with an initial velocity of +7.7 m/s. What will the ball's velocity be when it returns to starting point?

Answer in units of m/s. How long will the ball take to reach its starting point, and answer in units of s?

Asked on by kela000

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tonys538's profile pic

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

A tennis ball is thrown vertically upward with an initial velocity of +7.7 m/s.

As the ball rises, its velocity decreases at a rate equal to 9.8 m/s^2 which is the gravitational acceleration.

At the highest point, the velocity of the ball is 0. It then starts to fall down and reaches the ground in the same duration of time it took to rise to the highest point. The velocity of the ball is equal in magnitude to the velocity it was thrown up but opposite in direction.

This gives the velocity of the ball when it returns to the starting point as -7.7 m/s

The time taken by the ball is 2*(7.7/9.8) = 1.57 seconds

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

s = ut +(1/2) gt^2      (1),is the equation of motion of the tennis ball.

s is the vertical displacement at any time from the starting point, and u is the initial velocity of the tennis ball(=7.7m/s) and  g is the acceleration due to gravity(=9.8m/s^2) and t is the time of motion

The ball when reaches back at the starting point its net displacement is 0. So, using this in (1) we get the time to reach back the starting point.

0=7.7t-(1/2)(9.8)t^2

0=15.4t+0.8t^2

(t)(15.4-9.8t) = 0

t=0 or t=15.4/9.8 =1.5714 secs is the time for the ball to reach back to the starting point. t=0 is the starting time when s=0.

The ball takes 7.7/g secs to loose its velocity to zero , which is 0.7857 secs  which is half the total time duration from start to reach back to the same postion.

So, from 0 velocity at the highest point with acceleration due to gravity, g m/s^2 falling free back in 0.7857 secs it regains the velocity = u+gt =0+gt= 9.8*0.7857 = 7.7 m/s downward while reaching the ground.(See its starting velocity was equal in magnitude but upward in direction.)

 

 

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

When the ball is thrown up it slows down as it goes up till it reaches a velocity = 0. At this time the ball starts to again fall down with increasing velocity in the opposite direction. In it's downward travel when it just crosses the starting point, its velocity is exactly equal to the original starting velocity, but in opposite direction.

Thus since the starting velocity of the ball u = 7.7 m/s

The final velocity when crossing the starting point during its fall will be v = -7.7 m/s.

the time 't' taken for the ball to accelerate from initial velocity to final velocity is given by the formula:

t = (v -u)/a

Where a = acceleration. As the ball is moving under the influence of earth's gravitation we know this to be = 9.81 m/(s^2).

Therefore: t = (v - u)/a = [7.7 -(-7.7)]/9.81 = 15.4/9.81 = 1.57 s (approximately)

Answer:

Ball's velocity when it returns to starting point will be -7.7 m/s

The ball will take 1.57 seconds to reach its starting point.

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