A projectile is launched from ground level to the top of a cliff which is 195m away and 155m high. If the projectile lands on top of the cliff 7.6s after it is fired, find the initial velocity of the projectile (mag. and direction). Neglect air resistance.

Refer to the figure shown below.

u = initial launch velocity, m/s

θ = launch angle, north of east, deg.

Wind resistance is ignored.

g = 9.8 m/s².

The horizontal component of the launch velocity is u cosθ.

Because the horizontal distance traveled in 7.6 s is 195 m, therefore

7.6 * (u cosθ) = 195

u cosθ = 25.658 (1)

The vertical component of the launch velocity is u sinθ.

Because the vertical height traveled in 7.6 s is 155 m, therefore

(u sinθ) * 7.6 - 0.5*9.8 * (7.6²) = 155

u sin θ - 37.24 = 20.395

usin θ = 57.635 (2)

From (1) and (2), obtain

tan θ = 57.635/25.658 = 2.246

θ = 66°

From (2), obtain

u = 57.635/sin (66°) = 63.09 m/s

Answer:

Velocity = 63.09 m/s

Direction = 66° north of east (or 66° measured counterclockwise from the x-axis)

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