A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 17.0 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.)
h(t) = 1.50 + 17.0t - 4.9t^2
Now just find t when h(t) = 0
To find the time the ball is in the air before it hits the ground, we can use the equations of motion.
The equation we need to use is the one that relates the final velocity (vf), initial velocity (vi), acceleration (a), and time (t):
vf = vi + at
In this case, since the ball is thrown straight up, the final velocity when it hits the ground will be -17.0 m/s (negative because the velocity is directed downwards), the initial velocity is +17.0 m/s, and the acceleration due to gravity is -9.8 m/s².
Plugging in the values into the equation, we get:
-17.0 m/s = 17.0 m/s + (-9.8 m/s²) * t
Simplifying the equation, we have:
-34.0 m/s = -9.8 m/s² * t
Dividing both sides by -9.8 m/s², we find:
t = 34.0 m/s / 9.8 m/s²
Calculating this, we get:
t ≈ 3.47 seconds
Therefore, the ball is in the air for approximately 3.47 seconds before it hits the ground.