Question

A object is launched into the air vertically from ground level and hits the ground after 7 seconds. What was the maximum height of the object?

, Hint: Use one of the 5 Equations of Motion below:de: 1/8 1
Max Height = ut + -at? t Version 2 Understanding that, at the max height: Velocity = 0 and Acceleration = 9.8

Answer 1

The maximum height of an object launched vertically and hitting the ground after 7 seconds is calculated using the equations of motions. The initial velocity was determined to be 34.3 m/s, and the maximum height reached was found to be 59.85 meters.

Step-by-step explanation:

Calculating the Maximum Height of a Launched Object

To calculate the maximum height reached by an object launched vertically and hitting the ground after 7 seconds, we need to use the equations of motion. Since the velocity at the maximum height is zero and the acceleration is due to gravity (9.8 m/s2), the following equation of motion can be used:

s = ut + ½ at2

But the time (t) to reach the maximum height would be half of the total time since the motion is symmetrical going up and coming down. Therefore, the time to reach the maximum height is 3.5 seconds (half of 7 seconds).

The initial velocity (u) can be determined by using the equation:

v = u + at

Solving for u gives us:

u = v - at

Since v is zero at the maximum height and the acceleration (a) is -9.8 m/s2, we can calculate u

u = 0 - (-9.8 m/s2) × 3.5 s

u = 34.3 m/s

We then use the initial velocity to determine the maximum height (s) by plugging u back into the first equation:

s = (34.3 m/s) × 3.5 s + ½ (-9.8 m/s2) × (3.5 s)2

After calculating, we find that the maximum height is 59.85 meters.