A projectile is shot upward from the surface of the Earth with an initial velocity of 120 meters per second. What is its velocity after 5 seconds?

Your answers are wrong. 71m/s

To find the velocity of the projectile after 5 seconds, we need to use the equations of motion for vertical motion with constant acceleration.

The given initial velocity is 120 m/s. As the projectile is shot upward against gravity, the acceleration will be negative due to gravity pulling it downward. Taking the acceleration due to gravity as -9.8 m/s^2 (assuming no air resistance), we can use the following equation to find the velocity after 5 seconds:

Final velocity (v) = Initial velocity (u) + (acceleration (a) * time (t))

Plugging in the values:

v = 120 m/s + (-9.8 m/s^2) * 5 s

Calculating:

v = 120 m/s + (-49 m/s)

v = 71 m/s

Therefore, the velocity of the projectile after 5 seconds is 71 meters per second.

Height = -4.9t^2 + 120t

d(height)/dt = -9.8t +120

when t=5
velocity = -9.8(25)+120 = -120 m/s

Height = -4.9t^2 + 120t

d(height)/dt = -9.8t +120

when t=5
velocity = -9.8(25)+120 = -120 m/s