Two students are on a balcony a distance h above the S street. One student throws a ball vertically downward at a speed vi ; at the same time, the other student throws a ball vertically upward at the same speed. Answer the following symbolically in terms of vi, g, h, and t. (a) What is the time interval between when the first ball strikes the ground and the second ball strikes the ground? (b) Find the velocity of each ball as it strikes the ground. (c) How far apart are the balls at a time t after they are thrown and before they strike
ground?
Any answers??? Any help???
Someone please help
To answer these questions, we can use the equations of motion for uniformly accelerated motion. Let's go through each part step by step.
(a) What is the time interval between when the first ball strikes the ground and the second ball strikes the ground?
To find the time interval, we need to calculate the time it takes for the first ball to hit the ground and subtract it from the time it takes for the second ball to hit the ground.
For the first ball, we know that its initial velocity (vi) is downward, and it is accelerating due to gravity (g) in the downward direction. The equation we can use is:
h = (vi * t) + (1/2)g * t^2
Since the ball hits the ground, we can set h to zero and solve for t:
0 = (vi * t) + (1/2)g * t^2
This is a quadratic equation, and we can use the quadratic formula to solve for t. Once we have the time it takes for the first ball to hit the ground, let's call it t1.
For the second ball, the situation is the same, but the initial velocity (vi) is upward. So we have:
0 = (-vi * t) + (1/2)g * t^2
Again, solving for t gives us the time it takes for the second ball to hit the ground, let's call it t2.
Then, the time interval between the two events is t2 - t1.
(b) Find the velocity of each ball as it strikes the ground.
To find the velocity of each ball as it strikes the ground, we can use the equation of motion:
v = vi + g * t
For the first ball, substituting the time t1, we can find its velocity v1 as it strikes the ground.
For the second ball, substituting the time t2, we can find its velocity v2 as it strikes the ground.
(c) How far apart are the balls at a time t after they are thrown and before they strike the ground?
To find the distance between the balls at a time t, we need to determine their positions at that time first.
For the first ball, its position at time t is given by:
h1 = (vi * t) + (1/2)g * t^2
And for the second ball, its position at time t is given by:
h2 = (-vi * t) + (1/2)g * t^2
The distance between the balls at time t is then h2 - h1.
Solving these without numbers is an godly mess. Who assigned this??
a)the two times are solved via quadratics:
0 = 1/2 g t^2 + vi t - h and
0 = 1/2 g t^2 - vi t - h
So t1 = [-vi + sqrt(vi^2 - 4(1/2 g)(-h)]/2(1/2g)
which is ridiculous to work with
b) the velocities will be the same:
vf = vi + at
c) x = 1/2 (vf + vi) t