Question

1: A baseball is hit 4 feet above the ground leaves the bat with an initial speed of 98 ft/sec at an angle of 0 45 is caught by an outfielder at a height of 3 feet.

Answer 1

299.36 feet

Step-by-step explanation:

`To \ find \ the \ distance \ of \ the \ ball \ from \ the \ home \ plate. \\ \\ From \ the \ given \ information:`

`Height \ h = 4 \ ft`

`Initial \ speed \ V_o = 98 \ ft/s ec`

`The \ angle \ \theta = 45^0`

`Acceleration \ due \ to \ gravity (g)= 32.2 \ ft/s`

`U_x = V_o \ cos 45 = (98)/(√(2))`

`U_y = V_o \ sin 45 = (98)/(√(2))`

So;

`S_y = u_y t - (1)/(2)gt^2`

`-1 =(98)/(√(2))t - (1)/(2)*32*1.85t^2`

By solving:

`t_1 = 4.32 \ sec`

Thus;

`horizontal \ distance = U_x t`

`= (98)/(√(2))* 4.32`

`\mathbf{=299.36 \ feet}`

`\mathbf{Thus \ , the \ distance \ from \ the \ home \ plate \ = \ 299.36 \ feet}`