The figure shows the paths of a golf ball your friend drops from the window of her apartment and of the rock you throw from the ground at the same instant. The rock and the ball collide at x = 41.00 m, y = 17.02 m, and t = 2.94 s. If the ball was dropped from a height of h = 59.42 m, determine
a) the initial velocity of the rock,
b) the velocity of the rock at the time of the collision with the ball.
a. h = Yo*t + 4.9g*t^2 = 17.02 m
Yo*2.94 - 4.9*(2.94)^2 = 17.02
2.94Yo - 42.35 = 17.02
2.94Yo = 17.02 + 42.35 = 59.37
Yo = 20.2 m/s = Ver. component of initial velocity.
Dx = Xo*2.94 = 41
Xo = 13.95 m/s = Hor. component of
initial velocity.
tanA = Yo/Xo = 20.2/13.95 = 1.44803
A = 55.4o
Vo = 13.95/cos55.4 = 24.6 m/s[55.4o].
b. Y^2 = Yo^2 + 2g*h
Y^2 = 20.2^2 - 19.6*17.02 = 74.4
Y = 8.63 m/s.
tanA = Y/Xo = 8.63/13.95 = 0.61864
A = 31.74o
V = 13.95/cos31.74 = 16.4m/s[31.74o].
To determine the initial velocity of the rock and its velocity at the time of the collision with the ball, we can use the principles of projectile motion and kinematics.
Let's break down the problem into two parts:
a) Determining the initial velocity of the rock:
1. We know that the ball was dropped from a height of 59.42 m. We can use this information to find the time it took for the ball to fall to the point of collision.
2. We can use the kinematic equation for vertical displacement to calculate the time it took for the ball to fall. The equation is given by:
Δy = v0y * t + (1/2) * g * t^2
Where:
Δy is the vertical displacement (59.42 m)
v0y is the initial vertical velocity of the ball (which is zero since it was dropped, so v0y = 0)
t is the time it took for the ball to fall
g is the acceleration due to gravity (9.8 m/s^2)
3. Rearranging the equation, we get:
Δy = (1/2) * g * t^2
substituting the values, we get:
59.42 m = (1/2) * 9.8 m/s^2 * t^2
Simplifying, we get:
2 * 59.42 m = 9.8 m/s^2 * t^2
118.84 m = 9.8 m/s^2 * t^2
Solving for t, we get:
t^2 = 12.12 s^2
Taking the square root, we get:
t = 3.48 s (approx)
4. Now that we have the time it took for the ball to fall, we can use this to determine the initial velocity of the rock. Since both the ball and the rock collide at the same time, the time is 2.94 s.
We can use the horizontal distance traveled by the rock to calculate the initial velocity using the formula:
Δx = v0x * t
Where:
Δx is the horizontal distance traveled by the rock (41.00 m)
v0x is the initial horizontal velocity of the rock (which we want to find)
t is the time it took for the rock to travel the distance (2.94 s)
Rearranging the equation, we get:
v0x = Δx / t
Substituting the values, we get:
v0x = 41.00 m / 2.94 s
Calculating, we find:
v0x = 13.95 m/s
Therefore, the initial horizontal velocity of the rock is approximately 13.95 m/s.
b) Determining the velocity of the rock at the time of collision:
Now that we know the initial velocity of the rock, we can use the kinematic equation to find its final velocity.
The kinematic equation for vertical motion is:
vf = v0y + gt
Where:
vf is the final vertical velocity of the rock
v0y is the initial vertical velocity of the rock (which is zero)
g is the acceleration due to gravity (9.8 m/s^2)
t is the time of collision (2.94 s)
As the rock is thrown horizontally, the vertical velocity remains constant.
Therefore,
vf = 0 + 9.8 m/s^2 * 2.94 s
Calculating, we get:
vf = 28.73 m/s
Therefore, the velocity of the rock at the time of the collision with the ball is approximately 28.73 m/s.