Question

A rock is launched from the top of a building 50 m high at an
angle of 30^{o} with a speed of 20 m/s. (a) How far from
the building does the rock land? (b) What is the rock's maximum
altitude as measured from the ground? (c) What is the velocity of
the rock just before landing?

Answer #1

Height of the building, h = 50 m

Angle of projection, θ =
30^{0}

Speed of projection, u = 20 m/s

Let, acceleration due to gravity, g
= 10 m/s^{2}

The equation of motion of this projectile is

-h = (u Sin θ) T – ½ g
T^{2} where T is the total time for the flight.

- 50 = (20 x Sin 30 ) xT – ½ x 10 x
T^{2}

T^{2} – 2 T + 10 =
0

Solving the equation, we get, T = 4.32 s

(a)

Range of the rock, R = u Cos θ x T = 20 x Cos 30 x 4.32 = 74.8m

The rock will land at a distance 78.4 m from the building.

(b)

The maximum altitude of the rock, H
= h + (u Sin θ)^{2}/2g = 50 + (20 x Sin 30 )^{2} /
2x 10 = 55 m

The maximum altitude of the rock, H = 55 m

( c )

The velocity of rock just before
landing, V = (u^{2} + 2gh)^{1/2} = (20^{2}
+ 2 x 10 x 50)^{1/2} = 37.4 m/s

The velocity of rock just before landing, V = 37.4 m/s

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