A projectile is thrown from the top of a building with an initial velocity of 30 m/s in the horizontal direction. If the top of the building is 30 m above the ground, how fast will the projectile be moving just before it strikes the ground?
A) 35 m/s
B) 39 m/s
C) 31 m/s
D) 43 m/s
E) 54 m/s

Question

Quizplus · Accepted Answer

Since the vertical velocity of the projectile is affected by gravity, we can use the kinematic equation:

y = viy*t + 0.5*a*t^2

Where y is the vertical displacement (30 m), viy is the initial vertical velocity (0 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time the projectile takes to reach the ground.

Solving for t, we get:

t = sqrt(2*y/a) = sqrt(2*30/9.8) = 2.19 s

Now, we can use the horizontal velocity to find the horizontal displacement of the projectile in this time:

x = vix*t = 30*2.19 = 65.7 m

Finally, we can use the Pythagorean theorem to find the magnitude of the final velocity:

v = sqrt(vix^2 + viy^2 + 2*a*y) = sqrt(30^2 + 0 + 2*9.8*30) = 39.2 m/s

Rounding to the nearest integer, the answer is B: 39 m/s.

A Projectile Is Thrown from the Top of a Building