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 8.26 s. What is its initial velocity? Neglect air resistance.
Its vertical distance is given by a parabola.
Since we know the t-intercepts are 0 and 8.26 ,and we know a =-4.9 m/sec^2
the equation is
s = -4.9t(t-8.26)
= -4.9t^2 + 40.472 , where m is in metres
v = -9.8t + 40.472
when t = 0 , v = 40.472
initial velocity is 40.72 m/sec
Wow, that is quite a toss!
BTW, if you are working in feet,
a = -32 ft/sec^2
duplicate the above steps, not much has to be changed.
To solve this problem, we can use the equations of motion for vertically thrown objects and the given time of flight.
In this case, the motion of the ball can be divided into two parts: the upward motion and the downward motion. Since we know that the ball returns to its point of release, the times taken for the upward and downward motions are equal. Therefore, each motion takes half of the total time, which is 8.26 s divided by 2, equal to 4.13 s.
Let's denote the initial velocity of the ball as "v0." When the ball reaches its maximum height, its vertical velocity becomes 0, and we can use this fact to determine the initial velocity.
Using the equation of motion relating initial velocity, final velocity, acceleration, and time:
v = u + at
where:
v = final velocity (0 m/s when the ball reaches maximum height)
u = initial velocity (what we want to find)
a = acceleration (acceleration due to gravity, -9.8 m/s^2, since it acts in the opposite direction of the positive direction)
t = time (4.13 s, each for upward and downward motion)
Plugging in the values, we get:
0 = v0 + (-9.8 m/s^2) * 4.13 s
Simplifying the equation:
-9.8 m/s^2 * 4.13 s = v0
Solving for v0:
v0 = 40.474 m/s
Therefore, the initial velocity of the ball when it was thrown vertically upward is approximately 40.474 m/s.