Question

A stone is thrown vertically up with an intial velocity 49ms^(-1) from the top of a tower and reaches ground after 12 seconds.Find the height of the tower if you ans I will follow you​

Answer 1

Given:

• Initial velocity :- 49 ms^(-1)
• Time :- 12 seconds
• According due to gravity, acceleration :- 9.8 m/sq-sec

To Find:

Height of tower.

Solution:

Given that, Initial velocity :- 49ms^(-1)

and time :- 12sec.

Concept:

☞ According to the question, a stone is thrown vertically up with an given initial velocity from the top of a tower and reaches ground after sometime.

☞ Newton developed three equation of motion are,

• v = u + at
• s = ut + 1/2 at^2
• v^2 = u^2 - 2as

Now,

According to the question, we used here 2nd equation of motion.

Calculation:

➡ s = ut + 1/2 at^2

➡ 49 × 12 - (1/2) 9.8×12×12

➡ 49 × 12 - (9.8×12×12/2)

➡ 49 × 12 (1-1/2)

➡ 49 × 12 × (-0.2)

➡ -117.6 m

•°• The height of tower is -117.6 m.

Answer 2

Answer:

Height of the tower = 117.6m

Explanation:

Given that,

Initial velocity = u = 49m/s

Time taken to go upwards and then reach downwards = 12s

Acceleration due to gravity = a = g = 9.8m/s²

First let us look at the case where the stone is throw up and reaches maximum height point

Initial velocity = u = 49m/s

Final velocity = v = 0 m/s  (Body comes to rest at max height)

Acceleration due to gravity = a = -g = -9.8m/s²

Hence using v = u + at we get

0 = 49 - 9.8t

9.8t = 49

t = 5s

Therefore,

the stone takes (5 + 5) = 10s to go upwards and then reach the point where it was thrown

Hence,

stone takes (12 - 10) = 2s to reach the bottom of the tower

In the second case, where the ball moves down with a velocity

Initial velocity = u = 49m/s

Time taken = 2s

Acceleration = a = g = 9.8m/s²

Hence using S = ut + 1/2at² we get,

S = 49×2 + 1/2 × 9.8 × 2²

S = 98 + 19.6

S = 117.6 m