Suppose the ball is thrown from the same height as in the practice it problem at an angle of 39.0° below the horizontal. if it strikes the ground 41.8 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 flight s (b) the initial speed m/s (c) the speed and angle of the velocity vector with respect to the horizontal at impact speed m/s angle ° below the horizontal

The data about the previous problem are missing: I found them online and attached here (see picture). The relevant missing data is just the height of the building, which is 45 meters.

(a) 1.51 s

The motion along the x-direction is uniform (constant speed), so the position along the x-direction is given by

where is the initial speed, is the angle of launching, t is the time. We can re-write the formula as

(1)

Along the y-direction the motion is accelerated (free-fall), so the equation for the vertical position is

(2)

where h = 45.0 m is the height of the building and is the acceleration of gravity. Substituting (1) into (2),

The ball strikes the ground when y(t)=0, so the equation becomes

And when this happens, the displacement along the x-axis is x = 41.8 m. Solving the equation for t and substituting the numbers, we find the time of flight:

(b) 35.6 m/s

We can now easily find the initial speed by looking at the equation for the horizontal displacement:

Re-arranging it,

And substituting:

x = 41.8 m

t = 1.51 s

we find

(c) 46.4 m/s at degrees below the horizontal

The horizontal component of the velocity during the whole motion is constant, so it is

The vertical velocity instead follows the equation

where the negative signs means the direction is downward. Substituting t=1.51 s, we find the vertical velocity at the time of impact:

Therefore, the speed at the moment of impact is:

And the direction will be given by

the answer would be c. motion and direction

i hope this : )

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