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
