Question

A projectile is launched at an angle of 30° from the horizontal and lands 69 s later at the same height from which it was launched. (a) What is the initial speed of the projectile (in m/s)?

Answer 1

Answer:

679.2 m/s

Explanation:

The formula for the time of flight of a projectile is given as,

T = 2usinФ/g ......................... Equation 1

Where T = Time of flight, u = initial velocity, Ф = angle of projectile to the horizontal, g = acceleration due to gravity.

make u the subject of the equation

u = Tg/2sinФ....................... Equation 2

Given: T = 69 s, Ф = 30°

Constant: g = 9.8 m/s²

Substitute into equation 2

u = (69×9.8)/(2×sin30)

u = 679.2 m/s.

Hence the initial speed of the projectile is 679.2 m/s