Question

An arrow is fired downward at an angle of 45 degrees from the top of a 200 m cliff with a velocity of 60.0 m/s.

Answer 1

Answer:

A) 3.39 seconds

Explanation:

A) How long will it take the arrow to hit the ground?

We can find the time it takes the arrow to hit the ground by using a constant acceleration equation.

Let's make the positive direction upwards and the negative direction downwards. Let's list out the relevant variables:

• v₀ = -60 m/s
• Δy = -200 m
• a = -9.8 m/s²
• t = ?

Find a constant acceleration equation that contains these four variables.

• Δy = v₀t + 1/2at²

Substitute known values into the formula and solve for t.

• -200 = (-60 · cos(45))t + 1/2(-9.8)t²
• -200 = (-30√2)t - 4.9t²
• 0 = -4.9t² - (30√2)t + 200

Use the quadratic formula to continue solving for t.

• $\displaystyle \frac{-b \pm \sqrt{b^2-4ac} }{2a}$
• $\displaystyle \frac{-(-30\sqrt{2})\pm \sqrt{(-30\sqrt{2})^2 -4(-4.9)(200)} }{2(-4.9)}$
• $\displaystyle \frac{30\sqrt{2} \pm \sqrt{1800+3920} }{-9.8}$
• $\displaystyle \frac{30\sqrt{2} \pm \sqrt{5720} }{-9.8}$

From this, when we add the discriminant we get -12.04664168. When we subtract the discriminant we get 3.38819129.  

Since time cannot be negative, the time it takes for the arrow to hit the ground must be 3.39 seconds.

Answer 2

Answer:V^2=U^2+2gSIN45S

60^2=2asin45x200

3600=400sin45(a)

3600=282•84a

a=3600/282•84

a=12m/s^2

S=ut+0•5gt^2

200=0•5(12)t^2

200=6t^2

t^2=200/6

t^2=33.33

t=5.77s

Explanation: