A)...If this pitch were thrown horizontally, the ball would fall .809m by the time it reached home plate, 18.3m away. The acceleration of gravity is 9.81 m/s^2
How fast was the pitch
Answer in m/a
B)... A spy and an official reach the edge of a 5.4m waterfall. The spy's speed is 15 m/s and the officials' speed is 24 m/s.
How far apart will the two vessels be when they land below the waterfall?
The acceleration of gravity is 9.8 m/s^2
Answer in units of m
(A) L=18.3 m,
h=0.809 m
h=gt²/2
t=sqrt(2h/g) = ....
L=vt =>
v=L/t=....
(B)
h=gt²/2
t=sqrt(2h/g) = …
L₁=v₁t =...
L₂=v₂t =....
Thanks but I'm still confused about the formulas could you work it for me so I can better understand how to do it?
A) To find the initial speed of the pitch, we can use the equation of motion for a horizontally thrown object:
d = v0*t + (1/2)*a*t^2
In this case, we know the initial position (d = 0), the acceleration (a = -9.81 m/s^2), the final position (d = 18.3 m), and the time (t = ?). Rearranging the equation, we get:
18.3 = v0*t + (1/2)*(-9.81)*t^2
Since the ball was thrown horizontally, the initial vertical velocity (v0) is 0. So the equation becomes:
18.3 = (1/2)*(-9.81)*t^2
Simplifying:
-9.81*t^2 = 18.3
Solving for t:
t^2 = 18.3 / -9.81
t ≈ -1.883 seconds (ignore the negative solution, as time cannot be negative)
Now, we can use the time to find the initial horizontal velocity (v0). The horizontal distance traveled is given by:
d = v0*t
18.3 = v0 * (-1.883)
Solving for v0:
v0 ≈ -9.698 m/s (ignore the negative sign, as the velocity is in the positive direction)
Therefore, the speed of the pitch is approximately 9.698 m/s.
B) Both the spy and the official will be affected by the same acceleration due to gravity, so their vertical motion will be the same. The only difference will be in their horizontal motion.
The time taken for both vessels to reach the bottom of the waterfall can be found using the equation:
d = v0*t + (1/2)*a*t^2
For the spy, d = 5.4 m, v0 = 15 m/s, and a = 9.8 m/s^2. Solving for t:
5.4 = 15*t + (1/2)*9.8*t^2
Rearranging and simplifying:
4.9*t^2 + 15*t - 5.4 = 0
Using the quadratic formula, we get two solutions:
t ≈ 0.512 s and t ≈ -2.510 s (ignore the negative solution)
For the official, d = 5.4 m, v0 = 24 m/s, and a = 9.8 m/s^2. Solving for t:
5.4 = 24*t + (1/2)*9.8*t^2
Rearranging and simplifying:
4.9*t^2 + 24*t - 5.4 = 0
Again using the quadratic formula, we get two solutions:
t ≈ 0.202 s and t ≈ -2.767 s (ignore the negative solution)
Taking the longer time of 0.512 s (since both vessels will reach the bottom at the same time), we can use it to find the horizontal distance covered by both vessels.
For the spy:
d_spy = v0*t
d_spy = 15 * 0.512
d_spy ≈ 7.68 m
For the official:
d_official = v0*t
d_official = 24 * 0.512
d_official ≈ 12.29 m
Therefore, the two vessels will be approximately 7.68 m apart when they land below the waterfall.
A) To find the speed of the pitch, we can use the equation of motion for an object in free fall:
𝑑 = 𝑣0𝑡 + 0.5𝑎𝑡^2
Rearranging the equation to solve for the initial velocity, 𝑣0:
𝑣0 = (𝑑 - 0.5𝑎𝑡^2) / 𝑡
Given that the initial position, 𝑑, is 0 (since the pitch starts at the same height as the home plate), the acceleration, 𝑎, is -9.81 m/s^2 (negative because it's acting against the motion), and the time, 𝑡, is the time it takes for the pitch to reach home plate, we can substitute the values:
𝑣0 = (18.3 - 0.5 * 9.81 * 𝑡^2) / 𝑡
To find 𝑡, we can use the second equation of motion:
𝑑 = 𝑣0𝑡 + 0.5𝑎𝑡^2
Since the pitch falls 0.809m, we have:
0.809 = 𝑣0𝑡 + 0.5 * 9.81 * 𝑡^2
We now have two equations with two unknowns (𝑣0 and 𝑡). We can solve them simultaneously to find the speed of the pitch.
B) To find the distance between the two vessels when they land below the waterfall, we need to find the time it takes for each vessel to reach the bottom. Since both the spy and the official are dropped from the same height, the time of free fall will be the same for both.
We can use the equation of motion:
𝑑 = 𝑣0𝑡 + 0.5𝑎𝑡^2
Rearranging the equation to solve for time, 𝑡:
𝑡 = √((2𝑑) / 𝑎)
Given that both vessels fall from a height of 5.4m and the acceleration is -9.8 m/s^2 (negative because it's acting against the motion), we substitute the values:
𝑡 = √((2 * 5.4) / 9.8)
To find the distance between the two vessels, we can subtract the distances traveled by each vessel:
Distance = (vessel_speed) * time
Given that the spy's speed is 15 m/s and the official's speed is 24 m/s, we can calculate the distance:
Distance = (24 * 𝑡) - (15 * 𝑡)
Simplifying the expression gives us the final answer.