Take the acceleration due to gravity as g= 9.81ms-2
A red ball is thrown straight up from the edge of the roof of a building. A green ball is dropped from the same point (t = 1.80s) seconds later.
(a) If the height of the building is (h = 74) m, what is the initial velocity (ured) of the red ball, if both balls are to hit the ground at the same time ?
Now consider the same situation, but let the initial velocity be a known quantity and the height of the building (h) be the unknown. Take your value of t.
(b) Perform a calculation to determine the height of the building which would satisfy the condition that both balls would hit the ground at the same moment, for the following values of ured
(i) 12.0 ms-1 (ii) 19.1 ms-1
(c) for your value of t, if ured is greater than some maximum value umax there is no value of h that allows both balls to strike the ground at the same time.
(i) Determine the value of umax
(ii) There is a physical interpretation of this situation. What is it?
(d) for your value of t, if ured is less than some minimum value umin there is no value of h that allows both balls to strike the ground at the same time.
(i) Determine the value of umin
(ii) There is a physical interpretation of this situation. What is it?
what class is this lesson for?
(a) To find the initial velocity (ured) of the red ball, we can use the equations of motion for vertical motion.
For the red ball:
Initial velocity (ur) = ?
Acceleration due to gravity (g) = 9.81 m/s^2
Time taken (t) = ?
Since the red ball is thrown straight up, its final velocity when it reaches the topmost point (at height h) will be zero. We can use the equation:
vf = ur - gt
At the maximum height, the final velocity vf is zero, so we have:
0 = ur - gt
Rearranging this equation, we get:
ur = gt
Substituting the values of g = 9.81 m/s^2 and t = 1.8 s (given in the question), we can calculate the ur for the red ball.
(b) To calculate the height of the building (h) for different values of ured, we can use the equations of motion again.
For the green ball (dropped ball):
Initial velocity (ug) = 0 m/s
Acceleration due to gravity (g) = 9.81 m/s^2
Time taken (t) = 1.8 s
We can use the equation:
h = ug*t + (1/2)gt^2
Substituting ug = 0 m/s, g = 9.81 m/s^2, and t = 1.8 s, we can calculate the height of the building (h) for each of the given values of ured.
(i) ured = 12.0 m/s
(ii) ured = 19.1 m/s
(c) To determine the maximum initial velocity (umax) for which there is no value of h that allows both balls to hit the ground at the same time, we need to consider the case where the green ball is dropped at t = 0 s.
For the green ball:
Initial velocity (ug) = 0 m/s
Acceleration due to gravity (g) = 9.81 m/s^2
Time taken (t) = 1.8 s (given in the question)
To calculate umax, we need to find the time at which the red ball reaches the ground if it is thrown straight up with the maximum initial velocity. We can use the equation:
h = ur*t - (1/2)gt^2
Substituting ur = umax, g = 9.81 m/s^2, and t = 1.8 s, we can solve for h. If the value of h is less than 74 m (the height of the building), then there is no value of h that allows both balls to hit the ground at the same time.
(ii) The physical interpretation of this situation is that the maximum initial velocity (umax) represents the maximum velocity at which the red ball can be thrown straight up and still hit the ground, given the height of the building.
(d) To determine the minimum initial velocity (umin) for which there is no value of h that allows both balls to hit the ground at the same time, we need to consider the case where the green ball is dropped at t = 0 s.
For the green ball:
Initial velocity (ug) = 0 m/s
Acceleration due to gravity (g) = 9.81 m/s^2
Time taken (t) = 1.8 s (given in the question)
To calculate umin, we need to find the time at which the red ball reaches the ground if it is thrown straight up with the minimum initial velocity. We can use the equation:
h = ur*t - (1/2)gt^2
Substituting ur = umin, g = 9.81 m/s^2, and t = 1.8 s, we can solve for h. If the value of h is greater than 74 m (the height of the building), then there is no value of h that allows both balls to hit the ground at the same time.
(ii) The physical interpretation of this situation is that the minimum initial velocity (umin) represents the minimum velocity at which the red ball needs to be thrown straight up in order to hit the ground, given the height of the building.