Question

a ball thrown vertically upwards with a velocity of 19.6 metre per second from the top of a tower returns to earth in 12 seconds calculate the height of the tower if g is equal to 9.8 metre per second square

Answer 1

Hi friend,

We know that formula,


V²-u²=2gh

Here the initial velocity is zero,

The velocity will be negative as it moving against the gravity.

(-19.6)²=2×(9.8)×H

H= 19.6×19.6÷2×9.8

Height= 19.6m.

Hope this helped you a little!!!

Answer 2

$\maltese \: \red{\mid{\underline{\overline{\textbf{Answer}}}\mid}}$

It is given that :-

• Returns to earth in = 12s
• Since thrown, Initial velocity= 0
• Final velocity = 19.6m
• Acceleration due to gravity = 9.8
• Height of tower = ?

Height of tower = Distance travelled by ball.

Now according to third equation of motion, square of velocity of a body is equals to sum of square of initial velocity and double of product of acceleration and distance. Represented by :-

$\maltese \: {v}^{2} = {u}^{2} + 2as$

In this equation :

• v = final velocity
• u = Initial velocity
• a = acceleration
• s = distance

Now after inputting known values in this equation we get :

$s = \frac{ {v}^{2} - {u}^{2} }{2a}$

$\maltese \: s = \frac{ {(19.6)}^{2} - 0 }{2 \times 9.8}$

$\maltese \: s = \frac{19.6 \times 19.6}{19.6}$

$\maltese \: s \: = \: \cancel \frac{19.6}{19.6} \times 19.6$

$\maltese \: s = 1 \times 19.6$

$\maltese \: s = 19.6m$

Therefore height of tower is 19.6m.

_______________________________________

BY SCIVIBHANSHU

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