Answer:
a. The final speed , v' when it returns is the same speed, u as when it was released. v' = u
b. The time of flight T equals twice the time taken to reach the highest point, t. T = 2t
Explanation:
(a) As long as you can neglect the effects of the air, if you throw anything vertically upward, it will have the same speed when it returns to the release point as when it was released.
Using v² = u² - 2gh, we find the maximum height the object reaches. With u = initial vertical velocity, v = velocity at maximum height = 0 m/s (since it momentarily stops), g = acceleration due to gravity and h = maximum height
So, v² = u² - 2gh
0² = u² - 2gh
0 = u² - 2gh
-u² = -2gh
h = -u²/-2g
h = u²/2g
On its return trip to the point where it is released, the distance covered is h and its initial velocity u' = 0
With v' as the final velocity, we use
v'² = u'² + 2gh
substituting the values of the variables into the equation, we have
v'² = 0² + 2g(u²/2g)
v'² = 0² + u²
v'² = u²
taking square-root of both sides, we have
√v'² = √u²
v' = u
So, the final speed , v' when it returns is the same speed u as when it was released. v' = u
(b) The time of flight will be twice the time it takes to get to its highest point.
At the highest point, its velocity is zero.
Using v = u - gt where u = initial velocity, v = final velocity(velocity at highest point) = 0 (since it is momentarily at rest), g = acceleration due to gravity and t = time taken to reach the highest point.
So, substituting the values of the variables into the equation, we have
v = u - gt
0 = u - gt
-u = -gt
t = u/g
On its return trip, since we know its final velocity = u and its initial velocity is 0 m/s, again, using
v' = u' + gt' where u' = initial velocity = 0 (since it is at its highest point), v' = final velocity = u (velocity on return to release point), g = acceleration due to gravity and t' = time taken to reach release point from highest point
So, substituting the values of the variables into the equation, we have
v' = u' + gt'
u = 0 + gt'
u = gt'
t' = u/g = t
So, the time of flight, T = time taken to reach highest point + time taken to return to release point
T = t + t'
T = u/g +u/g
T = 2u/g
T = 2t
So, the time of flight T equals twice the time taken to reach the highest point, t. T = 2t
