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. h = 0.5g*t^2 = 0.809 m.
4.9t^2 = 0.809
t^2 = 0.1651
t = 0.4063 s.
Xo * t = 18.3 m.
Xo * 0.4063 = 18.3
Xo = 45 m/s. = Velocity of the pitch.
B. h = 0.5g*t^2 = 5.4 m
4.9*t^2 = 5.4
t^2 = 1.1
Tf = = 1.05 = Fall time of each vessel.
Ds = Xo * Tf = 15m/s * 1.05s. = 15.75m.
= Dist. of the spy.
Do = 24m/s * 1.05s. = 25.2 m. = Dist. of official.
Do - Ds = 25.2 - 15.75 = 9.45 m. Apart.
A) To find the speed of the pitch, we can use the kinematic equation:
d = v₀t + (1/2)at²
In this case, the initial vertical velocity (v₀) is 0 since the pitch is thrown horizontally, the distance (d) is 0.809 m, the time (t) is unknown, and the acceleration (a) is -9.81 m/s² (negative because it is acting downwards due to gravity).
Rearranging the equation, we get:
0.809 = (1/2)(-9.81)t²
Multiply both sides by 2 to get rid of the fraction:
1.618 = -9.81t²
Divide both sides by -9.81 to solve for t²:
t² = 1.618 / -9.81
t² ≈ -0.1652
Since time cannot be negative in this context, we discard the negative solution. Taking the square root of the remaining positive solution, we get:
t ≈ √(0.1652) ≈ 0.406 s
Now that we know the time, we can find the horizontal speed using the equation:
v = d/t
v = 18.3 m / 0.406 s
v ≈ 45.08 m/s
So, the speed of the pitch when thrown horizontally is approximately 45.08 m/s.
B) To find the distance between the two vessels when they land below the waterfall, we need to calculate the time it takes for both of them to reach the ground.
Using the equation:
d = v₀t + (1/2)at²
For the spy: v₀ = 15 m/s, a = 9.8 m/s² (acting downwards), and d = 5.4 m (distance of the waterfall).
5.4 = 15t + (1/2)(9.8)t²
Rearranging and simplifying:
4.9t² + 15t - 5.4 = 0
Solving this quadratic equation, we find two solutions for t. However, we discard the negative solution since time cannot be negative:
t ≈ 0.309 s
Now using the same equation for the official: v₀ = 24 m/s, a = 9.8 m/s², and d = 5.4 m.
5.4 = 24t + (1/2)(9.8)t²
4.9t² + 24t - 5.4 = 0
Again, solving this quadratic equation, we get:
t ≈ 0.247 s
The time for the official to reach the ground is less than the time for the spy.
Now, to find the distance between the two vessels when they land below the waterfall, we use the equation:
distance = velocity × time
For the spy: distance = 15 m/s × 0.309 s ≈ 4.635 m
For the official: distance = 24 m/s × 0.247 s ≈ 5.928 m
Therefore, the two vessels will be approximately 4.635 m + 5.928 m = 10.563 m apart when they land below the waterfall.