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=(x(t))/(v_0 cos \theta)` (1)

The vertical displacement instead is given by

`y(t) = -v_0 sin \theta t - (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 - (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:

`(1)/(2)gt^2=-y -x tan \theta\\t=\sqrt{(2(-y-xtan \theta))/(g)}=\sqrt{(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 = (x)/(t cos \theta)=(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=√(v_x^2+v_y^2)=√((31.0)^2+(-35.4)^2)=47.1 m/s`

While the direction is:

`\theta=tan^(-1)((v_y)/(v_x))=tan^(-1)((-35.4)/(31.0))=-48.8^(\circ)`