Question

You throw a ball with a speed of 25.0 m/s at an angle of 40.0â—¦ above the horizontal directly toward a wall. The wall is 22.0 m from the release point of the ball. How long does the ball take to reach the wall?
Show Options
A.11.15 s
B.11.5 s
C.1.115 s
D.1.15 s

Answer 1

Answer:

D.1.15 s

Explanation:

When we decompose the velocity into two perpendicular components, the total vector and its components form a right triangle. Therefore, these components will be given by:

$cos\theta=\frac{v_x}{v}\\v_x=vcos\theta(1)\\sin\theta=\frac{v_y}{v}\\v_y=vsin\theta$

Using (1), we calculate the horizontal speed of the ball:

$v_x=25\frac{m}{s}cos(40^\circ)\\v_x=19.15\frac{m}{s}$

Recall that$v=\frac{x}{t}$, solving for t:

$t=\frac{x}{v}\\t=\frac{22m}{19.15\frac{m}{s}}\\t=1.15s$

Answer 2

The horizontal speed is going to be the cosine of the given speed, therefore, the horizontal speed is 19.15 m/s. To find the time, divide the 22 m distance by the velocity. This results in 1.131 seconds, which is in between C and D.