Question

A ball is projected upwards with velocity of

200 m/s. Find the speed of the ball at half the

maximum height. Take g = 10 m/s2.
​

Answer 1

Solution :
Initial speed= u=50m/s
g=10m/s2
At maximum height V=0 m/s

Maximum height:
V²-u²=2as
0²-50²=2*-10*s
s=50*50/20
s=125m


the speed at half the maximum height:
s=125/2=62.5m
v=?
u=50m/s
a=-10m/s²
From v²-u²=2as
V=√u²+2as
V=√50²+2x(-10)x 62.5
v=√2500-1250
v=√1250=35 m/s

Answer 2

Answer:

20 √ 5 m / sec .

Explanation:

Given :

Initial velocity ( u )  = 200 m / sec

Acceleration due to gravity ( g ) = 10 m / sec

First finding height :

H = u² / 2 g

H = 40000 / 2 × 10 m

H = 2000 m

But we have to find velocity at H / 2 .

Using equation of motion under gravity :

v² = u² - 2 g H / 2

Putting values here we get

v² = 40000 - 2 × 10 × H / 2

v² = 40000 - 20000 m / sec

v² = 2000

v = √ 2000 m / sec

v = 20 √ 5 m / sec .

Hence we get answer .