100 points! a ball is thrown 19.0 m/s at an angle of 40.0º with the horizontal. assume the ball is thrown at ground level 22.0 meters away. a. how long does it take the ball to reach the wall? b. at what height does the ball hit the wall?

A. 1.51 seconds

B. 7.26 meters

Explanation:

A. Time to reach 22 meters

Find horizontal component of initial velocity V = 19 m/s:

Vx= Vcos(theta) = 19cos(40)  = 14.55 m/s

Use equation d= V0t + (1/2)at^2 to find the time

(acceleration is 0 for horizontal component) so: d = Vx*t

t = d/Vx = 22/14.55 = 1.51 seconds

B. Find vertical component of initial velocity:

Vy= Vsin(theta) = 19sin(40) = 12.21

Use equation y = V0*t + (1/2)at^2 to find the height at 1.51 seconds

y = (12.21)(1.51) + (.5)(-9.81)(1.51)^2

y= 7.26 meters

answer(20) psi;