Answer to Question #175082 in Mechanics | Relativity for Eren Yeager

Question

A person throws a baseball with a horizontal component of velocity of 25 m/s. After 2 seconds, the ball is 40 m. above the release point. Calculate the horizontal distance it has traveled by this time, it’s initial vertical component of velocity, and it’s initial angle of projection. Also, is the ball on the way up or the way down at this moment in time?

Answer 1

Solution.

$v_{0x}=25m/s;$

$t=2s;$

$h=40m;$

$L-?; v_{0y}-? \alpha-?;$

$L=v_{0x}t;$

$L=25m/s\sdot2s=50m;$

$h=v_{0y}t-\dfrac{gt^2}{2}\implies v_{0y}=\dfrac{(h+\dfrac{gt^2}{2})}{t};$

$v_{0y}=\dfrac{40m+20m}{2s}=30m/s;$

$tg\alpha=\dfrac{v_{0y}}{v_{0x}};$

$tg\alpha=\dfrac{25}{30}=\dfrac{5}{6};$

$\alpha=40^o;$

The ball moves up.

Answer: $L=50m; v_{0y}=30m/s;\alpha=40^o;$ the ball moves up.