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# physics problem ?

A ball is thrown straight upward and returns to the thrower’s hand after 1.81 s in the air. A second ball is thrown at an angle of 39.0 with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically? The acceleration of gravity is 9.81 m/s^2. Answer in units of m/s.

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- AshLv 71 month agoFavorite Answer
For 1st ball

Time of flight = 2v₀/g

1.81 = 2v₀/9.81

v₀ = (1.81*9.81)/2 ....(1)

The 2nd ball should have this same vertical velocity. If initial velocity is u₀

vertical velocity = u₀sin(39.0)

plug in (1)

u₀sin(39.0) = (1.81*9.81)/2

u₀ = (1.81*9.81)/(2sin(39.0))

u₀ = 14.1 m/s <------------velocity at which 2nd ball should be thrown

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