Question

A ball is thrown vertically upwards from the top of a tower with an initial velocity of 19.6 m/s. The ball reaches the ground after 5 s. Calculate: (i) the height of the tower, (ii) the velocity of the ball on reaching the ground. Take g=9.8 m/s^2.

Answer 1

1.  The height of the tower is 24.5 m.

2.  The velocity of the ball on reaching the ground is 29.4 m/s.

Explanation:

Given:

          Initial velovity of ball 'u' = 19.6 m/s

          Time taken by ball to reach ground 't' = 5 sec.

          Gravity 'g' = 9.8 m/s

Solution:

          we know that,

                                   ⇒ g = $\frac{v}{t}$  

                                   ⇒   t = $\frac{v}{g}$

                                          = $\frac{19.6}{9.8}$

So,                               ⇒   t = 2 sec.

it means time taken to reach max. height is 2 seconds.

It means 2 seconds are taken by the ball to reach back to tower and rest of 1 second to reach to ground.

it means first two seconds are taken to make final velocity 'v' is zero (max. height).

                                   ⇒ v = 0 m/s

Time taken is 1 second to calculate height of tower (written above).

So, using equation of motion,

                                   ⇒ $s = ut + \frac{1}{2} a t^{2}$

                                   ⇒ s = 19.6 x 1 + $\frac{1}{2}$ x 9.8 x $1^{2}$

                                   ⇒ s = 24.5 m

From zero initial velocity at time 2 seconds to ground where time is 5 seconds. Now, time taken is 3 seconds from max. height to ground.

                                         

now,                             ⇒ v = u + at

                                    ⇒ v = 0 + 9.8 x 3

                                    ⇒ v = 29.4 m/s

Answer 2

Answer:

Explanation:

1.  The height of the tower is 24.5 m.

2.  The velocity of the ball on reaching the ground is 29.4 m/s.

Explanation:

Given:

         Initial velovity of ball 'u' = 19.6 m/s

         Time taken by ball to reach ground 't' = 5 sec.

         Gravity 'g' = 9.8 m/s

Solution:

         we know that,

                                  ⇒ g = \frac{v}{t}  

                                  ⇒   t = \frac{v}{g}

                                         = \frac{19.6}{9.8}

So,                               ⇒   t = 2 sec.

it means time taken to reach max. height is 2 seconds.

It means 2 seconds are taken by the ball to reach back to tower and rest of 1 second to reach to ground.

it means first two seconds are taken to make final velocity 'v' is zero (max. height).

                                  ⇒ v = 0 m/s

Time taken is 1 second to calculate height of tower (written above).

So, using equation of motion,

                                  ⇒ s = ut + \frac{1}{2} a t^{2}

                                  ⇒ s = 19.6 x 1 + \frac{1}{2} x 9.8 x 1^{2}

                                  ⇒ s = 24.5 m

From zero initial velocity at time 2 seconds to ground where time is 5 seconds. Now, time taken is 3 seconds from max. height to ground.

                                       

now,                             ⇒ v = u + at

                                   ⇒ v = 0 + 9.8 x 3

                                   ⇒ v = 29.4 m/s

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