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        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.
10 months ago

Vikas TU
11143 Points
							Hiii V^2 = u ^2 + 2asv = 0 so , u^2 = 20H Now, apply this in middle , V = 10m/s and s = H/2 so height comes out to be 10m

10 months ago
sree lakshmi
42 Points
							Let us suppose that the ball was thrown with an initial speed uNow, let us assume the Maximum Height reached is HA 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.

10 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions