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

Answer 2

The final speed of the projectile just before it is about to strike the ground is 38.57 m/s

Finding the final speed:

Given that the initial speed of the projectile is u = 30 m/s.

The height of the building from the ground is 30m.

So if we take the top of the building as the origin, then the final position of the projectile which is the ground is at a distance s = -30m

Applying the third equation of motion:

v² = u² - 2gs

where v is the final speed

v² = 30² - 2×9.8×(-30)

v² = 1488 m²/s²

v = 38.57 m/s

Learn more about laws of motion:

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