When a particle is thrown vertically up it remains at a height 'h' for 5second, above height H for 3second. Then find H-h.
Answer (1)
13th Jul, 2020
When the particle is thrown , the initial velocity will be zero. So, using h = ut + 1/2gt^2, we get
h = 0 + 1/2*10*(5^2)
h = 125m
Now, the velocity after t = 5 seconds can be find using v^2 - u^2 = 2gh
So, v^2 = 2*10*125
v = 50 m/s
Now, this v is the u for the latter case, during 5th to 8th second.
So, H = 50*3 + 1/2*10*3*3
H = 195m
So, H-h = (195-125)m = 70m
Hope this helps.
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