Question

Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0°below the horizontal. If it strikes the ground 50.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)(a) the time of flights(b) the initial speedm/s(c) the speed and angle of the velocity vector with respect to the horizontal at impactspeed m/sangle ° below the horizontal

Answer 1

The figure of the problem is missing: find in attachment.

(a) 1.64 s

The ball follows a projectile motion path. The horizontal displacement is given by

$x(t) = v_0 cos \theta t$

where

$v_0$ is the initial speed

t is the time

$\theta=32.0^{\circ}$ is the angle below the horizontal

We can rewrite this equation as

$t=\frac{x(t)}{v_0 cos \theta}$ (1)

The vertical displacement instead is given by

$y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2$ (2)

where

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting (1) into (2),

$y(t) = -x(t) tan \theta - \frac{1}{2}gt^2$

We know that for t = time of flight, the horizontal displacement is

$x(t) =50.8 m$

We also know that the vertical displacement is

$y(t) = -45 m$

Substituting everything into the equation, we can find the time of flight:

$\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s$

(b) 36.5 m/s

We can now find the initial speed directly by using the equation for the horizontal displacement:

$x(t) = v_0 cos \theta t$

where we have

x = 50.8 m

$\theta=32.0^{\circ}$

Substituting the time of flight,

t = 1.64 s

We find:

$v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s$

(c) 47.1 m/s at 48.8 degrees below the horizontal

As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to

$v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s$

The initial vertical velocity is instead

$u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s$

And it changes according to the equation

$v_y = u_y -gt$

So at t = 1.64 s (when the ball hits the ground),

$v_y = -19.3 - (9.8)(1.64)=-35.4 m/s$

So the impact speed is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s$

While the direction is:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}$