Answer to Question #138402 in Classical Mechanics for Addmore Mugabe

Question

You throw a ball with a speed of 25.0 m/s at an angle of 40.0◦ above the horizontal directly toward a
wall, as shown in Fig. 3.3. The wall is 22.0m 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 horizontal and vertical components of its velocity as it hits the wall? (d) When it hits, has it
passed the highest point on its trajectory?

Answer 1

solution

(a)

horizontal component of velocity(v_x)=25cos40⁰=19.15m/s

so horizontal displacement in time t

s_x=u_x.t

so

$22=19.15t\\t=\frac{22}{19.15}=1.14s$

so time taken by ball to reach the wall is 1.14s.

(b)

displacement of ball in vertical direction in 1.14s

$s_y=u_yt-\frac{1}{2}gt^2$

so

$s_y=16.06\times1.14-\frac{1}{2}\times9.8\times1.14^2$

$s_y=11.94m$

so ball is 11.94 m above the ground when it hit the ball

(c)

there are no acceleration in horizontal direction so u_x remain constant.

so horizontal component is

$v_x=19.15 m/s$

and vertical component is

$v_y=u_y-gt\\=16.06-9.8\times1.14\\=4.89m/s$

(D)

time taken by ball to reach the highest point can be calculated as\

at highest point

v_y=0m/s

so

$0=u_y-gt_1\\t_1=\frac{16.06}{9.8}=1.63 s$

$t<t_1$

so when ball will hit the wall it has not reached the highest point of trajectory.