Question

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

Answer 1

Answer:

V = 38.582 m/s

Explanation:

Initial Vertical Velocity = 0

Let's use the motion equation given below to solve for the final vertical velocity:

$v^2 - u^2 = 2*a*s$

Here a = 9.81 m/s^2

and s = 30 m

Solving for v, we get:

v = 24.261 m/s

This is just the vertical velocity, to get the total velocity we need to add the horizontal component of velocity with this:

Total velocity = Sqrt ( Horizontal Velocity^2 + Vertical Velocity^2)

$V = \sqrt{(V_x)^2+(V_y)^2}$

where $V_x = 30$  and $V_y = 24.261$

V = 38.582 m/s