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-) 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

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

where

is the initial speed

t is the time

is the angle below the horizontal

We can rewrite this equation as

(1)

The vertical displacement instead is given by

(2)

where

is the acceleration of gravity

Substituting (1) into (2),

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

We also know that the vertical displacement is

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

(b) 36.5 m/s

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

where we have

x = 50.8 m

Substituting the time of flight,

t = 1.64 s

We find:

(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

The initial vertical velocity is instead

And it changes according to the equation

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

So the impact speed is:

While the direction is:

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explanation:

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