A ball is thrown straight upward and returns to the thrower’s hand after 1 s in

Question

A ball is thrown straight upward and returns
to the thrower’s hand after 1 s in the air. A
second ball is thrown at an angle of 30 degrees 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 9.8 m/s2 .
Answer in units of m/s.

Explanation / Answer

Let the velocity of first ball be v and that of second ball be V The second ball is thrown at 30(deg) from the horizontal this means V(vertical) = Vsin30 now it takes 1 sec for the ball to come down back to the ground we have s = ut +.5gt^2 s= 0 u =v g= -9.8 t= 1 0 = v*1 -.5*9.8*1^2 => v= 4.9m/sec For the other ball to reach the same height vertically we need the vertical velocity of the to be same of that of the older ball so the velocity of the other ball is =Vsin30 = 4.9m/sec => V= 4.9*2 = 9.8m/sec

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