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Grade: 9
        
A ball is thrown vertically upwards.When it is half of its maximum height its velocity is 10 metre per second ,find the maximum height reached by ball.
one month ago

Answers : (2)

Vikas TU
8478 Points
							
Hiii 
V^2 = u ^2 + 2as
v = 0 
so , u^2 = 20H 
Now, apply this in middle , V = 10m/s 
and s = H/2 
so height comes out to be 10m 
 
one month ago
sree lakshmi
38 Points
							
Let us suppose that the ball was thrown with an initial speed u
Now, let us assume the Maximum Height reached is H
A constant acceleration g=10m/s2  is acting in downward direction. 

Let us consider the motion from start to Max Height.
In this time interval, final velocity is zero, and distance travelled is H. We have:
v^{2} = u^{2}+2as
\rightarrow 0^{2}=u^{2}+2\times (-10)H
u^{2}=20H
 
Now, let us consider the time from start to when the ball reaches half the maximum height. Now, in this time interval, final velocity is 10 m/s, and Distance Travelled is H/2
v^{2} = u^{2}+2as
10^{2}=20H +2(-10)H/2
100=20H-10H
10H=100
H= 10m
 
Thus, maximum height reached is 10 metres.
one month ago
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