Question

Determine the launch speed of a horizontally launched projectile that lands 26.3m from the base of a 19.3m high cliff.

Answer 1

The launch velocity of the projectile is 13.28 m/s.

What is projectile motion?

The motion of an object thrown in the air under the force of gravity is known as projectile motion.

Since the object is launched horizontally, its initial velocity along the vertical direction is zero. From the second kinematic equation,

s=u*t+(1/2)at^2.

where s is the displacement, t is the time, u is the initial velocity and a is the acceleration. Since the height is decreasing, so it will be taken negative.

For the vertical motion, s=-19.3 m, a=-9.8 m/s^2 and u=0. Put the values in the above equation and solve it.

-19.3 = (0)*t+(1/2)*(-9.8)*t^2

19.3 = (1/2)*(9.8)*t^2

t=1.98 s

Since the velocity along the horizontal direction is constant, the displacement along the horizontal direction is given by the formula,

X=vt

where X is the horizontal displacement, v is the initial horizontal velocity and t is the time.

For the horizontal motion, X=26.3 m and t=1.98 s. Put the values in this equation and solve it.

26.3=v*(1.98)

v=13.28 m/s

The launch velocity is equal to the initial horizontal velocity, so it is equal to 13.28 m/s.

Learn more about projectile motion.

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