Question

A stone is dropped from the top of a tower 78.4 m high, find the time taken to reach the ground at the velocity with it strike the ground

Answer 1

Answer:

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Explanation:

Given

h=78.4m

a=9.8m/sec^2

u=0m/sec

t=?

v=?

From 3rd equation of motion

$<font color=blue>v^2=u^2+2as</font color>$

${v}^{2} = {0}^{2} + 2 \times 9.8 \times 78.4 \\ v = \sqrt{1536.64} \\ v = 39.2m {sec}^{ - 1}$

From 1st equation of motion

v=u+at

t=(v-u)÷a

t=39.2÷9.8

t=4 sec.

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Answer 2

The time taken to reach the ground is 4 seconds.

Explanation:

A stone is dropped from the top of a tower 78.4 m high. It is required to find the time taken to reach the ground at the velocity with it strike the ground. Initial velocity of the stone is 0. Using second equation of motion as :

$h=ut+\dfrac{1}{2}at^2$

Here, a = g and u = 0

$h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 78.4}{9.8}} \\\\t=4\ s$

So, the time taken to reach the ground is 4 seconds.

Learn more,

Kinematics

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