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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.
3 months ago

Vikas TU
9098 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

3 months ago
sree lakshmi
39 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.

3 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