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An object is thrown upward at a speed of 34.05 m/s. How long does the ball remain in the air?

User Hkon
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To find the time an object remains in the air after being thrown upwards, we must first identify the important variables.

The initial speed of the object - which is the speed at which the object is thrown - is 34.05 m/s.

Gravity is the force that pulls the object back to the ground and it is approximately 9.81 m/s² on Earth.

The first step is to determine the time it takes for the object to reach its highest point. As the object ascends, gravity slows it down until it comes to a stop at its highest point. This is when gravity overcomes the initial upward force. It's important to note that at the highest point, the object's speed is 0.

We calculate this time by dividing the initial speed by the gravitational force:

34.05 m/s ÷ 9.81 m/s² = 3.47 seconds

This is the time it takes for the object to reach the highest point of its trajectory.

The total time in the air is twice the time it takes to reach the highest point. This is because the time for the object to travel upwards and then back down to its starting height is symmetrical.

So we simply multiply the time it takes to reach maximum height by 2:

3.47 seconds * 2 = 6.94 seconds

Therefore, it will take approximately 6.94 seconds for the object to return to the ground after being thrown upwards at a speed of 34.05 m/s.

User TangoAlee
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