Physics, 18.02.2021 21:30 Daisy254

You are on the Pirates of the Caribbean attraction in the Magic Kingdom at Disney World. Your boat rides through a pirate battle, in which cannons on a ship and in a fort are firing at each other. While you are aware that the splashes in the water do not represent actual cannonballs, you begin to wonder about such battles in the days of the pirates. Sup-pose the fort and the ship are separated by 75.0 m. You see that the cannons in the fort are aimed so that their cannon-balls would be fired horizontally from a height of 7.00 m above the water. (a) You wonder at what speed they must be fired in order to hit the ship before falling in the water.
(b) Then, you think about the sludge that must build up inside the barrel of a cannon. This sludge should slow down the cannonballs. A question occurs in your mind: if the can-nonballs can be fired at only 50.0% of the speed found ear-lier, is it possible to fire them upward at some angle to the horizontal so that they would reach the ship?

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Answer from: jujurae03

a) v₀ₓ = 62.76 m / s, b) θ₁ = 17.6º, θ₂ = 67.0º

Explanation:

We can solve this exercise using the projectile launch ratios

a) Let's find the time it takes for the bullet to reach the water level

y = y₀ + v_{oy} t - ½ g t²

when it reaches the water its height is zero y = 0, as the bullet is fired horizontally its initial vertical velocity is zero

0 = y₀ + 0 - ½ g t²

t = $\sqrt{2y_o/g}$

t = $\sqrt{2 \ 7 /9.8}$

t = 1,195 s

now we can calculate the speed with the horizontal movement

x = v₀ₓ t

v₀ₓ = x / t

v₀ₓ = 75.0 / 1.195

v₀ₓ = 62.76 m / s

b) if the speed of the bullets is half of that found

v₀ = 62.76 / 2 = 31.38 m / s

let's write the expressions for the distance

x = v₀ cos θ t

y = y₀ + v_{oy} sin θ t - ½ g t²

t = $\frac{x}{v_o \ cos \theta}$

we substitute

$0 = y_o + v_o sin \theta \ \frac{x}{v_o \cos \thetay} - 1/2 g \ (\frac{x}{v_o \ cos \theta})^2$

$0 = y_o + x tan \theta - \frac{1}{2} g \ \frac{x^2}{ v_o^2 \ cos^2 \theta}$

let's use the identified trigonometry

sec² θ = 1 + tan² θ

sec θ = 1 / cos θ

we substitute

$0 = y_o + x tan \theta - \frac{g x^2}{2 v_o^2} ( 1 + tan^2 \theta)$

$\frac{g x^2}{2v_o^2} tan^2 \theta - x tan \theta + \frac{gx^2}{2v_o^2} - y_o = 0$

we change variable

tan θ = H

$\frac{gx^2}{2 v_o^2 } H^2 - x H + \frac{gx^2}{2v_o^2}-y_o =0$

we subtitle the values

$\frac{9.8 \ 75^2}{2 \ 31.38^2} H^2 - 75 H + \frac{9.8 \ 75^2}{2 \ 31.38^2}-7 =0$

27.99 H² - 75 H + 20.99 = 0

H² - 2.679 H + 0.75 = 0

we solve the quadratic equation

H = [2.679 ± $\sqrt{2.679^2 - 4 0.75}$ ] / 2

H = [2,679 ± 2,044] / 2

H₁ = 0.3175

H₂ = 2.3615

now we can find the angles

H₁ = tan θ₁

θ₁ = tan⁻¹ H₁

θ₁ = tan⁻¹ 0.3175

θ₁ = 17.6º

θ₂ = 67.0º

for these two angles the bullet hits the boat

Answer from: Quest

seeing the graph would

Answer from: Quest

If football was tossed in air without gravity or friction the ball would go into space

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