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?
Accepted Solution
A:
Answer:Step-by-step explanation:Let's split the analysis on two components, horizontal and vertical. Supposed no air resistence, the horizontal movement is given by the expression [tex] d=25.0 cos20° t[/tex]. Since it travels 50 m, solving for [tex]t[/tex] you get [tex]t=\frac2.0{cos20°} \approx 2 s[/tex].The vertical movement is given by the expression [tex] h=25.0sin20°t-\frac12gt^2[/tex], where [tex]g=9.81m/s^2[/tex] is the gravitational acceleration. The highest point is reached when the vertical velocity ([tex]v=25.0sin 20° -gt[/tex]) is zero, or at [tex]t=\frac{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. [/tex]Please note that the number are heavily approximated, do plug yours in a calculator