A particle is projected at an angle of 60 to the horizontal with a speed of 25m/s

A particle is projected at an angle of 60 to the horizontal with a speed of 25m/s

A particle is projected at an angle of 60 to the horizontal with a speed of 25m/s. Calculate: (a) time of flight of the particle, (b) horizontal component of the speed of the particle at its maximum height

Solution

∅ = 60°,
U = 25m/s
g = 10m/s²

(a) T = 2Usin∅/g
= 2x25sin60/10
T = 4.3sec

(B) U = Ucos∅
= 25cos∅
= 12.5m/s

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