Question

A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.

Using the formulas of the projectile motion, in how many seconds will the projectile strike the ground? (Round your answer to the nearest tenth of a second.)

Answer 1

After 26.28 seconds projectile returns 26.28 seconds.

Step-by-step explanation:

Initial velocity = 450 ft/s = 137.16 m/s

Angle, θ = 70°

Consider the vertical motion of projectile,

When the projectile return to the ground we have

Displacement, s = 0 m

Acceleration, a = -9.81 m/s²

Initial velocity, u = 137.16 x sin70 = 128.89 m/s

Substituting in s = ut + 0.5 at²

s = ut + 0.5 at²

0 = 128.89 x t + 0.5 x (-9.81) x t²

t² - 26.28 t = 0

t ( t- 26.28) = 0

t = 0 s or t = 26.28 s

After 26.28 seconds projectile returns 26.28 seconds.