Answer to Question #91697 in Classical Mechanics for Louie

Question

Question 11
A person standing on level beach throws a ball with a velocity of 22 m/s at an angle of 30o above the horizontal. How far away is the ball when it hits the beach and what is the time of flight?

Answer 1

$Vx=Vi\times Cos\theta$

This is the horizontal component of velocity

$Vx=22\times Cos30^0$

$Vx=19.05\frac{m}{s}$

Now the vertical component of velocity$Vy=Vi\times Sin\theta$

$Vy=22*Sin30^0$

$Vy=11\frac{m}{s}$

Time of flight

$t=\frac{2Vy}{a}$

t=2*11/9.81

$t=2.24sec$

Distance traveled when hit the beach again

D=Horizontal velocity multiply by time of flight

D=19.05x2.24

D=42.76meter