Question

Please i need help A baseball is thrown at an angle of 20º relative to the ground at a speed of 25.0 m/s. If the ball was caught 50.0 m from the thrower, how long was it in the air? (1 point)

How high did the baseball travel before beginning it's descent?

Answer 1

Explanation:

Let's split the analysis on two components, horizontal and vertical.

Supposed no air resistence, the horizontal movement is given by the expression
`d=25.0 cos20° t`. Since it travels 50 m, solving for
`t` you get
`t=\frac2.0{cos20°} \approx 2 s`.

The vertical movement is given by the expression
`h=25.0sin20°t-\frac12gt^2`, where
`g=9.81m/s^2` is the gravitational acceleration. The highest point is reached when the vertical velocity (
`v=25.0sin 20° -gt`) is zero, or at
`t=(25.0sin20°)/(9.81) \approx 1s. At this time, it's height will be [tex] h= 25.0sin20° (1) -\frac1/2 (9.81) (1^2) \approx 4 m.`

Please note that the number are heavily approximated, do plug yours in a calculator