Question

A 0.25 kg object is projected vertically into the air with a velocity of 45 m/s. How high above the ground is the object after 4.3 seconds?

Answer 1

After 4.3 seconds, the projected object with an initial vertical velocity of 45 m/s is 102.7 meters above the ground.

Step-by-step explanation:

To calculate how high the object is after 4.3 seconds, we can use the kinematic equation for vertical motion (s = ut + 0.5at^2) where s is the displacement, u is the initial velocity, a is the acceleration due to gravity (which is -9.81 m/s^2 since it acts downwards), and t is the time.

Using the initial velocity (u = 45 m/s) and the time (t = 4.3 s), we calculate:

s = (45 m/s * 4.3 s) + 0.5 * (-9.81 m/s^2) * (4.3 s)^2

This calculates to:

s = 193.5 m - 90.8 m

s = 102.7 m

Therefore, the object is 102.7 meters above the ground after 4.3 seconds.