A ball is thrown 83.3 m upward and then falls
back to earth.Neglecting air resistance, how long will it
be in the air? The acceleration of gravity is
9.8 m/s2 .
Answer in units of s
To find out how long the ball will be in the air, we can use kinematic equations.
First, we need to determine the time it takes for the ball to reach its highest point. We can use the second equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity at the highest point (0 m/s since the ball momentarily stops),
u = initial velocity (which is the velocity at which the ball was thrown),
a = acceleration (which is the acceleration due to gravity, -9.8 m/s^2),
s = displacement (which is 83.3 m upwards).
Rearranging the equation, we get:
u^2 = v^2 - 2as
Since v = 0, the equation simplifies to:
u^2 = -2as
Now, we can solve for the initial velocity (u):
u^2 = -2as
u^2 = -2 * (-9.8 m/s^2) * 83.3 m
u^2 = 1628.84 m^2/s^2
Taking the square root of both sides:
u = √(1628.84 m^2/s^2)
u ≈ 40.36 m/s
Now that we know the initial velocity of 40.36 m/s, we can determine the time it takes for the ball to reach the highest point using the first equation of motion:
v = u + at
Where:
v = final velocity (0 m/s),
u = initial velocity (40.36 m/s),
a = acceleration (-9.8 m/s^2),
t = time (unknown).
Rearranging the equation, we get:
t = (v - u) / a
t = (0 m/s - 40.36 m/s) / (-9.8 m/s^2)
t ≈ 4.13 s
It takes approximately 4.13 seconds for the ball to reach its highest point. Since the total time in the air will be twice the time taken to reach the highest point, the ball will be in the air for around 2 * 4.13 seconds.
Therefore, the ball will be in the air for approximately 8.26 seconds.
hf=hi+vi*t-4.9 t^2
hf, hi =0
solve for t.