Ball thrown at an angle reaches 8.0m height and 16m range

[rachael]

A boy throws a ball into the air at an angle to the horizontal. The ball reaches a vertical height of 8.0m and travel a horizontal distance of 16m before being at the same vertical position as at the point of release
A. what is the magnitude of the vertical velocity component of the ball as it is thrown?
B. What is the time of the flight of the ball?
C. What is the horizontal velocity component of the ball at its point of throwing?
d. What is the speed of the ball at this point of release?
Thanks

 

[Pengwuino]

You need to show some work or we can't help you.

 

[lightgrav]

Find the time to fall down from the top first. the time to rise is the same.
That's (B).
Then do (c) or (A).

 

[phucnv87]

A boy throws a ball into the air at an angle to the horizontal. The ball reaches a vertical height of 8.0m and travel a horizontal distance of 16m before being at the same vertical position as at the point of release
A. what is the magnitude of the vertical velocity component of the ball as it is thrown?
B. What is the time of the flight of the ball?
C. What is the horizontal velocity component of the ball at its point of throwing?
d. What is the speed of the ball at this point of release?
Thanks

A) The maximum height of the ball is $$H=\frac{v_{0y}^2}{2g}$$
$$\Longrightarrow v_{0y}=\sqrt{2gH}$$

B)The vertical component of the ball $$v_y=v_{0y}-gt$$

The ball reach the max.height when $$v_y=0\Longrightarrow t=\frac{v_{0y}}{g}$$

The time of the flight of tha ball is $$\Delta t=2t=\frac{v_{oy}}{g}$$

C) The horizontal velocity component of the ball at its throwing point is $$v_{0x}=\frac{L}{\Delta t}$$

D) The speed of the ball at release point is $$v_0=\sqrt{v_{0x}^2+v_{0y}^2}$$[/color]