Answer to Question #142429 in Mechanics | Relativity for Tafadzwa Mubariri

Question

You throw a ball with a speed of 25m/s at an angle of 40degrees above horizontal directly toward a wall. The wall is 22m from the release point of the ball. (a) How long does the ball take to reach the wall? (b) How far above the release point does the ball hit the wall? (c) What are the vertical and horizontal components of its velocity as it hits the wall? (d) When it hits, has it passed the highest point on its trajectory?

Answer 1

(a)

$speed = 25m/s , angle = 40^{0} , d = 22m$

find the components of initial velocity and acceleration.

$v_{ox}= 25cos40^{0} = 19.15m/s, a_{x}= 0$

$v_{oy} = 25sin40^{0} = 16.06m/s, a_{y} = -g = -9.8m/s^{2}$

time taken, $t= \frac{d}{v_{ox}}=\frac{22}{19.15}= 1.15s$

(b) $y = y_{0} +v_{0}+1/2a^{2}_{y}= (16.06)(1.15)+0-1/2(9.8)(1.15)= 12m$

(c) the horizontal components of the ball's velocity is,

$v_{x}= v_{ox}+a_{x}t= 19.15+0(1.15)=19.2m/s$

vertical components of the ball's velocity is,

$v_{y}= v_{oy}+a_{y}t=(16.06)-(9.8)(1.15)=4.80m/s$

(d) Since $v_{y}$ is greater than zero, when the ball hits the wall, it has not reached the highest point yet.