A ball is projected vertically upward with a speed of 50 m/s-2. then the speed at half of the maximum height is
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u = 50 m/s, g = –10 m/s2 when moving upward, v = 0 (at highest point).
a) S = (V^2-u^2)/2a = (0-〖50〗^2)/(2(-10)) = 125 m
maximum height reached = 125 m
b) t = (v – u)/a = (0 – 50)/–10 = 5 sec
c) s’ = 125/2 = 62.5 m, u = 50 m/s, a = –10 m/s2,
v2 – u2 = 2as
⇒ v = √(( u^2+2as)) = √(〖50〗^2+2(-10)(62.5) ) = 35 m/s.
a) S = (V^2-u^2)/2a = (0-〖50〗^2)/(2(-10)) = 125 m
maximum height reached = 125 m
b) t = (v – u)/a = (0 – 50)/–10 = 5 sec
c) s’ = 125/2 = 62.5 m, u = 50 m/s, a = –10 m/s2,
v2 – u2 = 2as
⇒ v = √(( u^2+2as)) = √(〖50〗^2+2(-10)(62.5) ) = 35 m/s.
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