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A student fires a cannonball horizontally with a speed of 39m/s from a height of
73m. Neglect drag.

What was the cannonball's initial horizontal speed?

What was the cannonball's initial vertical speed?

How long did the ball remain in the air?

How far from the base of the building will the ball land (measured along the
ground)?

Respuesta :

MathPhys

Explanation:

The initial horizontal speed is 39 m/s, and the initial vertical speed is 0 m/s.

v₀ₓ = 39 m/s

v₀ᵧ = 0 m/s

In the y direction:

Δy = 73 m

v₀ᵧ = 0 m/s

aᵧ = 9.8 m/s²

Find: t

Δy = v₀ᵧ t + ½ aᵧt²

73 = 0 + ½ (9.8) t²

t = 3.9 s

In the x direction:

v₀ₓ = 39 m/s

aₓ = 0 m/s²

t = 3.9 s

Find: Δx

Δx = v₀ₓ t + ½ aₓt²

Δx = (39) (3.9) + 0

Δx = 150 m

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