Question

A ball is thrown vertically from the ground.it crosses a point at the height of 25m twice ,at an interval of 4s.the ball was thrown with a velocity of...... .m/s
1)20 2)25. 3)30 4)35

Answer 1

Answer:

30m/s

Explanation:

We know,

s=25m

total time=4s

time taken to reach maximum height=2s

v= 0

g= $-10m/s^2$  [when an object is thrown upwards, the force acting on it is -ve]

Let the initial velocity be 'u'

v=u + gt

0=u + [(-10) × 2]

=u + (-20)

-u= -20

Therefore, 'u' = 20m/s

Now, we apply the Law of Conservation of Energy

h= 25m

u= 20m/s

v= ?

$\frac{1}{2} mu^2 + mgh = \frac{1}{2} mv^2$

$u^2 + 2gh =v^2$

$v^2 = 20^2 + (2*10*25)$

$v^2=900$

$v=30$m/s