A ball is thrown vertically upwards with a velocity of 2 m/s. It is seen at height h twice.
Find the sum of the times, measured in seconds from the moment the ball is thrown, at which it is seen at height h.
Take g=10m/s²
Neglect air resistance.
Hints:
The average of the times is equal to the time to reach the maximum height, due the the symmetry of the parabola.
So the sum of the times would be twice the time it takes to reach the maximum height, also equal to the time for the object to go up and fall back to the original elevation. So figure out whichever is easier for you.
To find the sum of the times at which the ball is seen at height h, we need to determine the time it takes for the ball to reach height h for the first time and the time it takes for the ball to reach height h for the second time.
Given:
Initial velocity (u) = 2 m/s
Acceleration due to gravity (g) = 10 m/s²
Height (h) = h
We can use the kinematic equation to find the time it takes for the ball to reach height h:
h = u*t + (1/2)*g*t²
For the first time the ball reaches height h, we can set the equation as:
h = 2*t + (1/2)*10*t²
To find the time, we can rearrange the equation to:
t² + 2t - (2h/5) = 0
Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Applying it to our equation:
t = (-2 ± √(2² - 4*(1)*(-2h/5))) / (2*(1))
t = (-2 ± √(4 + 8h/5)) / 2
t = -1 ± √(1 + 2h/5)
Since we are considering the time it takes for the ball to reach height h for the first time, we take the positive value:
t₁ = -1 + √(1 + 2h/5)
Now, for the second time the ball reaches height h, we can set up another equation using the same logic as before. However, this time the initial velocity will be in the opposite direction (downwards) with a magnitude of -2 m/s:
h = -2*t + (1/2)*10*t²
Using the same quadratic formula, we get:
t₂ = -1 - √(1 + 2h/5)
The sum of the times at which the ball is seen at height h (t₁ + t₂) can be calculated as:
t₁ + t₂ = (-1 + √(1 + 2h/5)) + (-1 - √(1 + 2h/5))
= -2
Therefore, the sum of the times is -2 seconds.