a ball is projected from top of tower with a velocity of 5m/s at an angle of 53 to horizontal. what is its speed when it is at a height of 0.45 m from the point of projection ??
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Answer:
4 m/s
Explanation:acc to law of conservation of energy
1/2mv^2=mgh+1/2mv^2
1/2m(5)^2=mg(0.45)+1/2mv^2
v^2=(5)^2-2g(0.45)
v^2=25-9
v=4m/s
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