A ball is thrown horizontally from the top of a tower 40 m high. The ball strikes the ground at a point 80 m from the bottom of the tower. Find the angle that the velocity vector makes with the horizontal just before the ball hits the gound.

Options

A. `45^(@)`
B. `90^(@)`
C. `37^(@)`
D. `53^(@)`

Answer

A

Explanation

With standard kinematics, we get

`y=1/2 g t^(2), t=sqrt(2y//g) =sqrt(80//10) =2.86 s `
`x=v_(x)t,v_(x)=x/t =80/2.86 =28.0 m//s`
`v_(y)^(2)-v_(0)^(2)=2a(y-y_(0))`
`vy=-sqrt(2gy) =-sqrt(2(9.8)(40))=28.0 m//s`
` theta=A tan ((v_(y))/(v_(x)))=315^(@)`