A ball is tossed so that it bounces off the ground, rises to height of 2 m, and then hits the ground

Question

again 0.75 m away from the first bounce.
a. How long is the ball in the air between the two bounces?
b. What is the velocity in the x-direction of the ball?
c. What is the speed of the ball just before the second bounce?
d. What is the angle of the velocity vector with respect to the ground right after its first bounce

Answer 1

a. To determine the time in air, we can use the formula for the height of an object in free fall:
h(t) = -(1/2)gt^2 + vt => 0 = -(1/2)gt^2 + vt - 2.
Here, h(t) is the height of the ball as a function of time, g is the gravitational acceleration (approximately 9.8 m/s^2), and v is the initial vertical velocity of the ball before the bounce.
We can use the quadratic formula to find t:
t = (-v ± √(v^2 - 4(1/2)(-2)))/(-g)
t = (-v ± √(v^2 + 4(1/2)(2)))/(g)

Since the ball starts at a height of 2 m, one solution to this equation will correspond to the initial launch of the ball, and the other will correspond to the second bounce. The second bounce is the one we want.
t1 = (-v + √(v^2 + 4(1/2)(2)))/(g)
t2 = (-v - √(v^2 + 4(1/2)(2)))/(g)
Since we know that t1 < 0 (it is the time when the ball was thrown) and we want t2 (the time when the ball hits the ground again), we can simply choose the positive root:
t2 = (-v + √(v^2 + 4(1/2)(2)))/(g)

Now we know that the time in the air is t2. The answer is approximately:
t2 ≈ 0.64 seconds.

b. To find the x-velocity of the ball, we can use the horizontal displacement of the ball: d = vt
t = 0.75 m
v_x = d/t = 0.75 m / 0.64 s = 1.17 m/s.

c. To find the speed of the ball just before the second bounce, we will first find the final y-velocity using the equation:
v_f_y = v_i_y + gt
The initial y-velocity, v_i_y, is the same as -v since we assumed that v_i_y is positive (upwards) and v_f_y is negative (downwards).
v_f_y = -v + gt
The speed of the ball just before the second bounce is given by the magnitude of the velocity vector, which can be found as the square root of the sum of the squares of the x- and y-components of the vector:
v_f = √(v_x^2 + v_f_y^2)
Using the values for v_x and t calculated earlier, we have:
v_f ≈ √(1.17^2 + (-v + 9.8*0.64)^2) ≈ 3.99 m/s.

d. The angle θ of the velocity vector with respect to the ground right after the first bounce can be found using the components of the vector and the tangent function:
tan(θ) = v_y/v_x
v_y is just the negative of the initial y-velocity (-v) we calculated earlier.
θ = arctan(v_y/v_x) = arctan((-v)/1.17)

Now, we don't know the exact value of v, but we can just give the angle in terms of v:
θ = arctan((-v)/1.17).

If we know the angle of the initial throw or other information, we could find an exact value for θ.