Question

A ball thrown vertically upwards with a speed u from the top of a tower reaches the ground in 9 seconds another ball thrown vertically downwards from the samw positiom with the same speed u tskes 4 seconds to reacj ground calculate the vakue of u

Answer 1

Given that,

Speed of first ball= u

Time of first ball= 9 sec

Time of second ball = 4 sec

For first case,

The ball is thrown from the top of tower  

So, height = -h

We need to calculate the distance

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Where, u = initial speed

t = time

g = acceleration due to gravity

$-h=u9-\dfrac{1}{2}\times9.8\times9^2$

$-h=9u-396.9$....(I)

For second case,

Another ball thrown vertically downwards,

Speed = -u

We need to calculate the distance

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value in to the formula

$-h=-u\times 4-\dfrac{1}{2}\times9.8\times4^2$

$-h=-4u-78.4$.....(II)

We need to calculate the value of u

Using equation (I) and equation (II)

$9u-396.9=-4u-78.4$

$9u+4u=-78.4+396.9$

$13u=318.5$

$u=\dfrac{318.5}{13}$

$u=24.5\ m/s$

Hence, The value of u is 24.5 m/s.