A stone thrown upward from the top of a tower 85m high reaches the ground in 5 s

(a) Calculate the greatest height above the ground

(b) The velocity with which it reaches the ground

(c) The time taken to reach the maximum height (g = 10ms^–2)

Consider a tower at the height of 85 m.

A ball is thrown upward.

Our discussion gets divided into 3 cases:

Case – 1: Ball goes up from tower to a max height

Case – 2: Ball comes down from the max height to tower

Case – 3: Ball goes down the tower to hit the ground

The formula to be used:

..1

..2

..3

Notations used:

v = Final Velocity; u = Initial Velocity; a = Acceleration; t = Time Taken; s = Distance Travelled

Note: Here Acceleration = Acceleration due to gravity

a = g If g acts in the direction of motion

a = –g If g acts in the direction opposite of motion

Case – 1: Ball moves upwards

Let Initial Velocity =

So,

Now, at max height, Final Velocity = 0. So, v = 0

Let distance travelled = s

Also, a = –g and time taken =

So, using equations of motion:

..4

Also by using equation 2, we get:

..5

Case – 2: Ball coming downwards

Here, u = 0

= ?

Now, using equation 3 here, we get:

..6

Now, finding v by using equation 2:

..7

From equation 4 and 6, we see that:

..8

So, the total time taken to go up and come back to the tower is given by.

Case – 3: Ball goes down the tower to hit the ground

Total time = 5 s

Time to Travel 85 m = 5 – = 5 – = 5 – ..9

Now, using equation 3,

(a) Greatest height above the ground = 85 +

= 85 +

= 85 +

= 88.2 m

(b) Velocity with which it reaches the ground

Here, u = = 8 ms^–1; a = g = 10 ms^–2;

t = 5 – = 5 – s

So, v = 8 + 10 × (5 – )

= 8 + 50 – 16

= 50 – 8

= 42 ms^–1

Hence, v = 42 ms^–1

(c) Time taken to reach the maximum height =

(From 4)

Now, s

So, the ime taken to reach the maximum height is 0.8s.