A projectile is thrown with an initial speed of 12 m/sec horizontally from the top of a building 64 meters above ground level.

Question

a)    Determine the time taken by the projectile to hit the ground.

b)    How far does the object land from the base of the building?

c)    What is the vertical velocity component of the projectile when it hits the ground?

d)    What is the horizontal velocity component of the projectile when it hits the ground?

e)    With what velocity (magnitude and direction) does it hit the ground?

Answer 1

This is a classic problem in Physics, and it can be solved by splitting the problem into the x part (which has no force, so the calculations are simple),

and the y part (which has the force of gravity, so it is accelerating the whole time).

Break this problem into two simpler problems:

(1) vx = vx0 = 12 m/s. Since there is no force in the x direction (there is no air resistance in this problem), vx will be 12 m/s forever.

That answers part d) vx = 12 m/s

(2) vy = a*t + vy0 = -9.8 m/s * t + 0 m/s, so vy = -9.8 m/s * t. We need to know t to answer parts (b), (c), and (e).

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(a) Determine the time taken by the projectile to hit the ground.

To find t, we know

sy = 0.5*a*t^2 + vy*t + sy0 = -4.9m/s^2 * t^2 + 0 m/s * t + 64 m,

and we want to know when sy = 0 m, we then have

0 m = -4.9m/s^2 * t^2 + 64 m, so

4.9m/s^2 * t^2 = 64 m, so

t^2 = 64 m / 4.9 m/s^2 = 13.06 s^2, so

t = sqrt(13.06 s^2)

t = 3.6 s, this is the answer to part (a). (rounded to 2 significant digits)

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(b) How far does the object land from the base of the building?

Since vx = 12 m/s always, and since sx = vx*t + sx0 = 12m/s * t + 0 m,

sx = 12m/s * t, and after t=3.6 s, we have

sx = 12m/s * 3.6s

sx = 43 m, this is the answer to part (b). (rounded to 2 significant digits)

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(c)  What is the vertical velocity component of the projectile when it hits the ground?

We know that vy = ay * t + vy0 = -9.8 m/s^2 + 0 m/s, so

vy = -9.8 m/s^2 * t. (vy will increase (negatively) at almost -10m/s every second)

It hits the ground at t = 3.6 s, so

vy = -9.8 m/s^2 * 3.6 s

vy = -35 m/s, this is the answer to part (c). (rounded to 2 significant digits)

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(d) What is the horizontal velocity component of the projectile when it hits the ground?

As we saw at the top of this problem,

vx = vx0 = 12 m/s, since there is no horizontal Force, there is no horizontal acceleration, and the velocity must remain constant

vx = 12 m/s, this is the answer to part (d)

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(e) With what velocity (magnitude and direction) does it hit the ground?

We know that vy = -35 m/s, and that vx = 12 m/s, so we have a triangle for the total velocity vector when the particle hits the ground (35.38 m/s down, 12 m/s to the right)

This gives us magnitude = sqrt(35 * 35 + 12 * 12) m/s

magnitude = sqrt(1369) m/s

magnitude = 37 m/s. (rounded to 2 significant digits)

AND gives us tan(theta) = vy / vx = -35 m/s / 12 m/s

tan(theta) = -2.9, so

theta = arctan(-2.9)

theta = -71 degrees (below the horizontal). (rounded to 2 significant digits)

magnitude of velocity is 37 m/s, and theta = -71 degrees below the horizon

this is the answer to part (e)

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