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

a. Find ta, the time that the arrow spends in the air.
b. 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?

Respuesta :

Answer:

a) The time that the arrow spends in the air = 6.7 s

b) 5.59 s

Explanation:

Range of a projectile is given as

R = u² sin 2θ/g

R = 220 m

u = initial velocity = ?

g = 9.8 m/s²

θ = 45°

220 = [u² sin (2×45°)]/9.8

u = 46.43 m/s

Time of travel for a projectile = (2u sin θ)/g = (2 × 46.43 × sin 45°)/9.8 = 6.7 s

b) Using the equations of motion, we need to first obtain the time of fall of the Apple.

u = initial velocity = 0 m/s (since the Apple is dropped)

g = 9.8 m/s²

t = ?

y = 6 m

y = ut + gt²/2

6 = 0 + (9.8t²)/2

4.9t² = 6

t = 1.11 s

The Apple takes 1.11s to reach the required height, hence the amount of time to wait after the arrow has been shot for it to meet the dropped Apple = 6.7 - 1.11 = 5.59 s

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