Sally Sue is atop a 10.0 m high platform where she will drop an apple straight down. On the ground
below, and 20.0 m (horizontally) away, Betty Lou will shoot an arrow at the apple. If Betty's arrow has a
release speed of 25.0 m/s, at what angle must she fire that arrow, and at what time must Sally drop the
apple (relative to the firing of the arrow), such that the arrow will strike the apple at the highest point in
the arrow's trajectory?
H = 10 m is the initial height of the apple.
x=20 m,
v(o)=25 m/s is initial velocity of the arrow
The angle at which the arrow starts is α .
h is the height at which the arrow strikes the apple (max height of the arrow).
We’ll use the expressions for the range, the height and the time of the projectile. You can find the derivation of these formulas in any textbook.
Horizontal x is the half of the projectile range
x=v(o)² •sin2α/2•g =>
sin2α =2•x•g/ v(o)² =2•20•9.8/25² =0.627,
2α =38.84º, α =19.42º.
The height of the projectile is
h= v(o)² •sin²α/2•g =625•0.11/2•9.8 =3.25 m.
The time that is needed for the arrow to rich the highest point is
t= v(o) •sinα/g=25•0.33/9.8 =0.84 s.
The apple coveres the vertical distance (H-h) for the time
t1 = sqrt(2•(H-h)/g) = sqrt(2(20-3.25)/9.8) = 1.84s.
Δt = t1-t=1.84 -0.84 = 1s.
Therefore, the arrow has to start its motion 1 s after the beginning of the apple motion.
I understand but you didn't put 10(initial height) into any of the equations?
To determine the angle at which Betty must fire the arrow and the timing at which Sally should drop the apple, we need to consider the horizontal and vertical components of motion.
Let's break down the problem step by step:
1. Find the time it takes for the arrow to reach the highest point in its trajectory:
- The arrow's vertical motion is affected by gravity, so we can use the kinematic equation: Δy = v₀yt + (1/2)at², where Δy is the displacement in the y-direction, v₀y is the initial vertical velocity, a is the acceleration due to gravity, and t is the time.
- Since the arrow reaches the highest point in its trajectory, its vertical velocity at that point will be zero. Therefore, v₀y = 0.
- The displacement in the y-direction is the height of the platform, h = 10.0 m. Substitute the known values into the equation:
10.0 m = (1/2)(-9.8 m/s²)t²
- Solve the equation for t. Rearrange the equation: t² = (2h)/g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
t² = (2 * 10.0 m)/(9.8 m/s²) = 2.04 s²
- Take the square root of both sides to find t: t ≈ 1.43 s
2. Determine the horizontal distance the apple falls in that time:
- The apple's horizontal motion is unaffected by gravity, so we can use the equation: Δx = v₀x * t, where Δx is the displacement in the x-direction, v₀x is the initial horizontal velocity, and t is the time.
- The initial horizontal velocity is the same as the release speed of the arrow, v₀x = 25.0 m/s.
- Substitute the known values into the equation:
Δx = 25.0 m/s * 1.43 s ≈ 35.75 m
3. Calculate the angle at which Betty must fire the arrow:
- The angle can be found using the trigonometric relationship: tanθ = opposite/adjacent. In this case, the opposite side is the height of the platform (10.0 m) and the adjacent side is the horizontal distance (35.75 m) at the time the apple is dropped.
- Rearrange the equation to solve for θ: θ = arctan(opposite/adjacent), where arctan is the inverse tangent function.
θ = arctan(10.0 m/35.75 m) ≈ 15.9°
Therefore, Betty Lou must fire the arrow at an angle of approximately 15.9° and Sally Sue should drop the apple at the same time Betty fires the arrow.