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.

A.how long will it take the arrow to hit the ground?
B.how far from the base

Answer 1

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

Step-by-step explanation:

Answer 2

A) 3.39 seconds

Step-by-step 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 (-b \pm √(b^2-4ac) )/(2a)`
• 
  `\displaystyle \frac{-(-30√(2))\pm \sqrt{(-30√(2))^2 -4(-4.9)(200)} }{2(-4.9)}`
• 
  `\displaystyle (30√(2) \pm √(1800+3920) )/(-9.8)`
• 
  `\displaystyle (30√(2) \pm √(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.