How long does the ball stay in the air before hitting the ground?

[bommer]

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 11.0 m/s when the hand is 1.50 m above the ground.
How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

 

[Andrew Mason]

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 11.0 m/s when the hand is 1.50 m above the ground
How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

Can you express y in terms of initial position, velocity, acceleration (g) and time? y = y0 + ___ +- _____

AM

 

[quicknote]

To determine the time the ball is in the air, we can use the equation for displacement (Δx) with initial velocity (v0), acceleration due to gravity (g), and time (t): Δx = v0t + 1/2gt^2.

In this scenario, the initial velocity is 11.0 m/s (positive since the ball is thrown upwards), the displacement is 1.50 m (since the student's hand is 1.50 m above the ground), and the acceleration due to gravity is -9.8 m/s^2 (negative since it is acting downwards).

Plugging in these values into the equation, we get: 1.50 m = (11.0 m/s)t + 1/2(-9.8 m/s^2)t^2.

Solving for t, we get two solutions: t = 0.2 seconds and t = 1.9 seconds. However, since the ball is thrown upwards and then falls back to the ground, we only need to consider the positive solution of t = 1.9 seconds.

Therefore, the ball is in the air for 1.9 seconds before it hits the ground.