What is t_a, the time that the arrow spends in the air?

Question

What is t_a, the time that the arrow spends in the air?

An arrow is shot at an angle of \theta = 45 degrees above the horizontal. The arrow hits a tree a horizontal distance D = 220 m away, at the same height above the ground as it was shot. Use g = 9.8 m/s^2 for the magnitude of the acceleration due to gravity.

What is t_a, the time that the arrow spends in the air?

How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree?

Answer 1

You'll have to work out the initial velocity, which you can do by knowing that the time taken to traverse the horizontal distance is equal to the time taken to travel vertically up and down.

The horizontal velocity is vCos45 and you know the horizontal distance. Get an expression for the time.

Get an expression for the time taken to reach maximum height (ie vertical velocity equals zero) and double it to allow for the trip back down.

Put these two expressions equal and you should be able to solve for v (initial velocity). You should get 44.3 m/s.

Plug this back into one of your expressions for time to get t = 7s.

Answer 2

Not enough information. You need to give the initial velocity.

For initial velocity = Vi, the vertical component is:
Vi sin45 = 0.71 Vi (all calcs should be to 2 sig figs)

Time to max altitude is:
t = v/g = 0.71 Vi / 9.8 = 0.072 Vi

Total time in flight is twice the above = 0.144 Vi

To calculate the time at which to drop the apple, you would need to know the height of the apple. For height = Ha:

t = (2Ha/9.8)^0.5 = 0.45 Ha^0.5

Now get those values and plug them in.

Answer 3

7 seconds