Question

A projectile is launched at a 30° angle relative to the ground. The projectile has an initial velocity of 15 m/s and travels through the air for 2 seconds.

What is the horizontal displacement (rounded to the nearest hundredth) in m?

Answer 1

25.98

Step-by-step explanation:

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Answer 2

The horizontal displacement of the projectile is 25.98 m

Step-by-step explanation:

Given;

angle of projection, θ = 30°

initial velocity of the projectile, v = 15 m/s

time of flight, t = 2 seconds

The horizontal displacement or range is given by;

R = vₓt

where;

vₓ is the horizontal component of the velocity

t is the time of flight

R = (15cos30)(2)

R = 25.981 m

R = 25.98 m (to the nearest hundredth)

Therefore, the horizontal displacement of the projectile is 25.98 m