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solong [7]
3 years ago
8

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
Physics
1 answer:
Svetlanka [38]3 years ago
3 0

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

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