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1 year ago

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A projectile is launched at an angle of 45° from the horizontal and lands 21 s later at the same height from which it was launch

ed. (a) What is the initial speed of the projectile (in m/s)? m/s (b) What is the maximum altitude (in m)? m (c) What is the range (in m)? m (d) Calculate the displacement (in m) from the point of launch to the position on its trajectory at 14 s. (Express your answer in vector form. Assume the projectile initially travels in the +x and +y-directions, where the +x-direction is horizontal and the +y-direction is straight up.) Δr = m

1 answer:

irinina [24]1 year ago

6 0

Answer:

a) initial speed of projectile = 145.5 m/s

b) Maximum altitude = 540 m

c) Range = 2160.6 m

d) r = (1440î + 480j) m

Explanation:

The distance at any time for the projectile is given by the relation - r² = x² + y²

where x = horizontal distance covered covered by the projectile and y = vertical distance coveredby the projectile

Let the initial velocity be u = ?

angle of projection be θ with respect to the horizontal = 45°

u = (uₓî + uᵧj) m/s

T = total time of flight = 21 s

t = any time during the flight of the projectile

a) Total time of flight = 2 uᵧ/g = (2u sin θ)/g

21 = (2u sin 45°)/9.8

u = 145.5 m/s

b) maximum altitude of the projectile = H

H = (u² sin² θ)/2g

H = (145.5² sin² 45°)/(2 × 9.8)

H = 540 m

c) According to projectile motion the maximum horizontal displacement is given by

x = R = uₓT = u cos(θ) T (since uₓ = u cos θ)

R = (145.5 cos 45°) × 21 = 2160.6 m

d) At 14 s,

x = uₓt = u cos(θ) t (since uₓ = u cos θ)

x = (145.5 cos 45°) × 14 = 1440.1 m

y = uᵧ t - 0.5gt² = [u sin(θ)] t - 0.5gt² = (145.5 sin 45°) × 14 - 0.5(9.8)(14) = 480 m

r = (1440î + 480j) m

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