How tall is the cliff if Bob threw the ball from exactly 5 ft above the cliff edge?

Question

Bob is a pitcher, and knows that the fastest he can throw the ball is 𝑣0=91.0mph. Bob starts the stopwatch as he throws the ball (with no way to measure the ball's initial trajectory) and watches carefully. The ball rises and then falls, and after 𝑡1=0.910s, the ball is once again level with Bob. Bob cannot see well enough to time when the ball hits the ground. Bob's friend then measures that the ball landed a distance of 𝑥=376ft from the base of the cliff.

The gravitational acceleration in imperial units is 32.2 ft/s^2.

How tall is the cliff if Bob threw the ball from exactly 5 ft above the cliff edge?

Answer 1

I'll start by writing out what we are given, converting everything to matching units, and defining variables we don't know yet:

v₀ = (91miles/hr)*(5280 feet/ 1 mile)*(1 hr/ 60 min)*(1 min/ 60 seconds) = 133.47 feet/second (initial velocity)

t₁ = 0.91 seconds (time until ball is level with Bob)

x = 376 feet (total distance)

g = 32.2 ft/s² (acceleration due to gravity)

t₂ = ? (total time until ball hits the ground)

v_x = ? (velocity in the x direction)

v_y = ? (velocity in the y direction)

θ = ? (upward angle that the ball is thrown)

y₀ = ? (the total vertical height at the starting point of the ball)

We are looking for the cliff height, h_cliff, which is related to our y₀ term:

h_cliff = y₀ - 5 ft

How do we find y_o? Let's use this equation for projectile motion, where y is the vertical position of the ball at time, t:

y = -gt²/2 + v₀sin(θ)t + y₀

Let's look at the point where the ball reaches the ground.

t = t₂

y = 0

Let's plug in what we know:

0 = (-32.2 ft/s² * t₂²)/2 + 133.47 ft/s * sin(θ)t₂ + y₀

We can't solve for y₀ until we find θ and t₂

Let's zoom in on the part of the diagram between t = 0 (starting point of the ball) and t₁ (when the ball is level with Bob) and redefine our coordinate system so that the point where Bob lets go of the ball is at (0,0). That means our y₀ = 0, and at t = t₁, the ball is back at y = 0. Let's look at that equation again:

y = -gt₁²/2 + v₀sin(θ)t₁ + y₀

Plug in what we know:

0 = (-32.2 ft/s²)*(0.91 s)²/2 + 133.47 ft/s * sin(θ) * 0.91 s + 0

Solve for θ:

0 = -13.33 ft + 121.46 ft * sin(θ)

sin (θ) = 0.11

θ = sin^-1(0.11) = 6.3°

Now, let's solve for t₂ using the following equation:

x = v₀cos(θ)t + x₀, where x is the horizontal distance travelled by the ball

376 ft = 133.47 ft/s * cos(6.3°) * t₂ + 0

t₂ = 2.834 seconds

Now we can solve for y₀ by plugging in θ and t₂:

0 = (-32.2 ft/s² * (2.834 s)²)/2 + 133.47 ft/s * sin(6.3°)*(2.834 s) + y₀

y₀ = 87.8 feet

Remember, y₀ is the total height of the ball at t=0. To find the height of the cliff, we subtract 5 feet

h_cliff = y₀ - 5 ft = 87.8 ft - 5 ft = 82.8 feet

Answer 2

y = h + 5 + v₀sinθt - gt²/2. If t = t₁ then y = h + 5 + v₀sinθt₁ - gt₁²/2 = h; 5 + 133.47·0.91sinθ - 16.1·0.91² = 0

91mi/h = 91·5280/3600 ft/s = 133.47 ft/s. sinθ = 0.0686; θ = 3.934°;

x = v₀cosθt; t = x/(v₀cosθ) = 376/(133.47·cos3.934°) = 2.824 s;

y = h + 5 + v₀sinθt - gt²/2 = 0; h = 16.1·2.824² - 133.47sin3.394°·2.824 - 5 = 97.54 ft