Question

a body is projected from the top of the tower of height 39.5 m with speed 9.8 ms per 1 .find the time to reach the ground​

Answer 1

Given :

• height of tower = 39.5 m

• initial velocity = 9.8 m/s

To find :

• time to reach the ground

Solution :

Distance travelled by the body is = height of tower = 39.5 m

initial velocity of body = 9.8 m/s

Now, first we will find final velocity.

We know that :-

$\underline{\boxed{\bf{ {v}^{2} - {u}^{2} = 2gs }}}$

Where :-

• v = final velocity
• u = initial velocity
• g = acceleration due to gravity ( 9.8 m/s²)
• s = distance

$\implies \sf{v}^{2} - {u}^{2} = 2gs$

$\implies \sf{v}^{2} - {(9.8)}^{2} = 2 \times 9.8 \times 39.5$

$\implies \sf{v}^{2} - 96.04 = 774.2$

$\implies \sf{v}^{2} = 774.2 + 96.04$

$\implies \sf{v}^{2} = 870.24$

$\implies \sf{v}^{2} = \sqrt{870.24}$

$\implies \sf{v}^{2} = \sqrt{ \dfrac{87024}{100} }$

$\implies \sf{v} = 2.9 \: (approx)$

therefore final velocity = 2.9 m/s

Now we will the following formula to find out time :-

$\underline{\boxed{\bf{ {v} = {u} + at }}}$

$\implies \sf \: 29.4 = 9.8+ 9.8t$

$\implies \sf \: 29.4 - 9.8= 9.8t$

$\implies \sf \: 19.6= 9.8t$

$\implies \sf \: \dfrac{19.6}{9.8} = t$

$\implies \sf \: 2 = t$

Answer :

Time taken by the body to reach the ground = 2 seconds ( approx)