A ball is thrown vertically upward, which is the positive direction. A little later it returns to its point of release. The ball is in the air for a total time of 7.99 s. What is its initial velocity? Neglect air resistance.
hf=hi+vi*t-1/2 g t^2
solve for vi
To find the initial velocity of the ball thrown vertically upward, we can use the equation of motion for vertical motion under constant acceleration.
The equation for displacement in vertical motion is given by:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity
t = time
a = acceleration (in this case, acceleration due to gravity, which is approximately 9.8 m/s^2)
In this problem, the ball is thrown vertically upward, so we need to consider the time it takes for the ball to reach its highest point and then return back to its point of release.
Let's assume the time taken for the ball to reach its highest point is t1, and the time taken for it to return back to its point of release is t2. Since the total time the ball is in the air is given as 7.99 seconds, we have:
t1 + t2 = 7.99 s
At its highest point, the ball momentarily comes to rest, which means the final velocity will be 0 m/s. Therefore, we can write the equation for the final velocity as:
v = u + at
As the ball is thrown vertically upward, in the opposite direction of gravity, the acceleration will be negative. Therefore, we have:
0 = u - 9.8t2
Rearranging this equation, we get:
u = 9.8t2
Now, let's consider the time taken to reach the highest point. The time taken to reach the highest point (t1) is half of the total time, as the ball spends an equal amount of time going up and coming down. Therefore, we have:
t1 = 7.99 / 2
t1 = 3.995 s
Using this value of t1, we can find t2:
t2 = 7.99 - t1
t2 = 7.99 - 3.995
t2 = 3.995 s
Substituting the value of t2 in the equation for initial velocity:
u = 9.8 * 3.995
u ≈ 39.01 m/s
Therefore, the initial velocity of the ball thrown vertically upward is approximately 39.01 m/s.