A ball is thrown vertically upward and comes back down. Total time of 7.95 s in air. What is its initial velocity? Neglect air resistance.
To find the initial velocity of the ball, we can use the fact that the total time of flight is the sum of the time it takes for the ball to reach its maximum height and the time it takes for the ball to fall back to its starting point.
Let's break down the problem into two parts:
1. Time to reach maximum height:
When the ball reaches its maximum height, its vertical velocity becomes zero. We can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is acceleration, and t is the time taken.
In this case, the acceleration is due to gravity and acts in the opposite direction of motion, so it will be negative (-9.8 m/s²). The velocity at the maximum height is 0 m/s. Assuming the upward direction as positive, we can rewrite the equation as:
0 = u - 9.8t₁.
2. Time to fall back:
The time it takes for the ball to fall back to its starting point is the same as the time it takes to reach the maximum height (t₁). As the ball falls, the acceleration due to gravity acts in the same direction as motion, so it will be positive (+9.8 m/s²). The equation to calculate the distance fallen (s) is given by: s = u₂t₂ + (½)gt₂², where u₂ is the final velocity, t₂ is the time, and g is the acceleration due to gravity.
Since the final velocity is the opposite of the initial velocity (-u), we can rewrite the equation as:
s = -ut₂ + (½)(-9.8)t₂².
Given that the total time of flight is 7.95 seconds, we can set up the equation:
7.95 = t₁ + t₂.
Now we have a system of equations:
0 = u - 9.8t₁,
s = -ut₂ + (½)(-9.8)t₂²,
7.95 = t₁ + t₂.
Neglecting air resistance, we can assume the displacement (s) is zero since the ball starts and ends at the same height.
By solving these equations simultaneously, we can find the value of u, which is the initial velocity of the ball.