# An Object Is Thrown Vertically Upward

A ball is thrown vertically upward with a speed of 26.3m/s. How high does it rise? How long does it take to reach its highest point?

How long does it take the ball to hit the ground after it reaches its highest point? What is its velocity when it returns to the level from which it started?

Answer the distance in m, time in s, and velocity in m/s.

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Initial velocity of the ball = 26.3m/s

At the highest point the ball reaches 0 velocity because of the pull of the acceleration due to gravity g.

So, the time taken to reach the highest point =(final velocity - initial velocity)/ acceleration due to gravity =(0-26.3)/ (-g)secs=2.6837 min. The direction of acceleration is opposite to velocity,which is away from the ground and taken as positive.

The height reached by the ball h= (v^2-u^2/(-2g) = (0^2-26.3^2)/(2*-9.8)=35.2903 m.

In the downward motionthe initial velocity u is 0 at the heighest pointthe acceleration is gand the final velocity is v towards earths= 35.2803m. So the equation of motion is:

v^2-u^2= 2gs or

v= sqrt(2gs) = sqrt(2*9.8*35.2903)=26.3m/s vertically down ward.

Time taken to reach the ground , t=(v-u)/g= (26.3-0)/9.8=2.6837 s

A ball is thrown vertically upward with a speed of 26.3 m/s.

As the ball rises its speed decreases at a rate equal to 9.8 m/s^2 which is the gravitational acceleration acting in the downward direction.

The ball rises till its speed becomes 0. If the ball rises for t seconds.

0 = 26.3 - 9.8*t

t = 26.3/9.8 = 2.68 seconds

The height of the ball after 2.68 seconds is 26.3*2.68 - 1/2*9.8*2.68^2 = 35.29 m

The time taken by the ball to return to the ground is also 2.68 seconds.

Its velocity when it returns is 26.3 m/s in the downward direction.

The answer given above is correct. However the problem can be solved using easier method.

The ball is thrown up with velocity u = 26.3 m/s

While travelling up the velocity of ball slows down till it reaches velocity v = 0 at its highest point.

The acceleration of ball due to gravitational pull a = 9.8 m/s^2

Thus total reduction in velocity when the ball reaches its highest point = v.

Therefore time taken to reach highest point t = v/a =26.3/9.8 = 2.6837 s (approximately)

The height to which ball rises s = t*(v + u)/2 = 2.6837*26.3/2 = 35.2093

We do not need to make separate calculations for the return path of the ball from highest point to the point to the starting level, as the velocities are exactly equal and opposite those of the upward journey.

Therefore velocity of ball when it returns to the starting level = -26.3 m/s.

Assuming starting point of the ball was the ground level, the time taken by ball to reach ground after it reaches the highest point = 2.6837 s