A ball was thrown straight up at 26.2 m/s then caught at the same place it was released. How many seconds was it in the air?
hf=hi+vivertical*t-4.9t^2
0=0+26.2t-4.9t^2 solve for time in air, t
To find the time the ball was in the air, we can use the equation of motion for vertical motion:
\[ v_f = v_i + at \]
where:
- \( v_f \) is the final velocity (which is 0 since the ball was caught at the same place it was released)
- \( v_i \) is the initial velocity (26.2 m/s)
- \( a \) is the acceleration (due to gravity, which is approximately \( -9.8 \, \text{m/s}^2 \))
- \( t \) is the time in seconds
Since the final velocity is 0, we can rewrite the equation as:
\[ 0 = 26.2 - 9.8t \]
Simplifying the equation:
\[ 9.8t = 26.2 \]
Dividing both sides of the equation by 9.8:
\[ t = \frac{26.2}{9.8} \]
Calculating this will give us the time the ball was in the air.
To determine how long the ball was in the air, we can use the equation of motion for an object in free fall.
When the ball is thrown straight up, it experiences a constant downward acceleration due to gravity. The initial velocity when the ball is thrown upwards is +26.2 m/s. At the highest point in its trajectory, the ball momentarily comes to rest and then accelerates downward with a rate of -9.8 m/s².
Since there are two phases to the ball's trajectory (going up and coming down), we can divide the total time into two halves: the time taken to reach the highest point, and the time taken to come back down.
First, let's find the time taken to reach the highest point (t1):
Using the equation of motion:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is time.
At the highest point, the final velocity becomes 0 m/s.
Therefore, 0 = 26.2 m/s - 9.8 m/s² * t1.
Rearranging the equation:
9.8 m/s² * t1 = 26.2 m/s,
t1 = 26.2 m/s / 9.8 m/s² ≈ 2.673 s.
Second, let's find the time taken to come back down (t2):
Using the same equation of motion, but with the acceleration now being -9.8 m/s² since the ball is accelerating downward:
0 = 0 m/s - 9.8 m/s² * t2.
Solving for t2:
t2 = 0 m/s / -9.8 m/s² ≈ 0 s.
Now, to find the total time of flight (T), we add the time taken to reach the highest point (t1) and the time taken to come back down (t2):
T = t1 + t2 = 2.673 s + 0 s = 2.673 s.
Therefore, the ball was in the air for approximately 2.673 seconds.