Respuesta :

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Answer:

using all projectile formulas respectively ...

A. time taken to reach maximum height = UsinTITA/g = 65 × sin 50 / 10

t = 65 × 0.7660 /10

t = 4.979 seconds

B . Total TIME OF FLIGHT = 2U sin♧/g

T = 2t = 2× 4.979 = 9.95seconds

C. Maximum Height attained = U^2sin^2 ♤/2g

H = (65)^2 × (0.7660)^2 / 2 × 10

H = 4225 × 0.5867 / 20

H = 123.9m

D. Horizontal Range = U^2 sin2tita/ g

R = 65)^2 × ( sin 50×2 ) / 10

R = 4225 × sin 100 / 10

R = 4160.81 / 10

R = 416m

Explanation:

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MathPhys

Explanation:

Given:

v₀ₓ = 65 m/s cos 50° = 41.78 m/s

aₓ = 0 m/s²

v₀ᵧ = 65 m/s sin 50° = 49.79 m/s

aᵧ = -9.8 m/s²

(a) Find t when vᵧ = 0 m/s.

vᵧ = aᵧ t + v₀ᵧ

0 m/s = (-9.8 m/s²) t + 49.79 m/s

t = 5.08 s

(b) Assuming the ball lands at the same height it was launched at, the total time in the air is double the time it takes to reach the maximum height.

t = 2 (5.08 s)

t = 10.2 s

(c) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2aᵧΔy

(0 m/s)² = (49.79 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 126 m

(d) Find Δx when t = 10.2 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (41.78 m/s) (10.2 s) + ½ (0 m/s²) (10.2 s)²

Δx = 425 m

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