Question

a body is projected vertically upwards with velocity u returns to starting point in 4 sec then find u

Answer 1

v=0 (max height)
a = 9.8 m/s2
t=4 s
t = u×u/2a
4 = u2 /2×9.8
u2 = 2×9.8×4
u = 8.8t

Answer 2

The initial velocity is 8.8m/s

Solution:

From the given,

initial velocity is u

The projected particle returns to the starting point with the time of 4s

Hence, we have to find the value of u

We know that

$a=9.8 \mathrm{m} / \mathrm{s}^{2}$

t=4 s

We have

$t=\frac{u^{2}}{2 a}$

Hence we get,

4=\frac{u^{2}}{2 \times 9.8}

Shift the number 4 to the right hand side and u^2 to the left hand side of the equation

Then,  

$u^{2}=2 \times 9.8 \times 4$

$u^{2}=78.4$

u = 8.8m/s  

Then, the value of intial velocity is 8.8 m/s.