Find ta. the time that the arrow spends in the air.

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/s2 for the
magnitude of the acceleration due to gravity.

Part A

Find ta, the time that the arrow spends in the air.

Part 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?

Answer

General guidance

Concepts and reason
The concepts related to solve this problem are horizontal range and time of flight of a projectile which is projected with some angle above the ground.

Initially, calculate the time that the arrow spends in the air. After that, calculate the time of descent. Finally, calculate the time taken for apple to fall from the tree.

Fundamentals

The expression for the horizontal range of a projectile which is projected with some angle above the ground can be calculated using the following formula:

D-V sin(20)

Here, is the initial velocity, g is the acceleration due to gravity, and is the angle of projection.

Rearrange the equation for.

V
=
Dg
V sin (20)

The time of flight of projectile which is projected with some angle above the ground can be calculated using the following formula:

1 - 2V sino

Step-by-step

Step 1 of 2

(a)

Substitute 220 m for D, 9.8 m/sfor g, and for in the equationV
=
Dg
V sin (20).

220 m(9.8 m/s)
sin (2(45°))
= 46.43 m/s

Substitute 46.43 m/s for V, 9.8 m/sfor g, and for in the equation1-2V, sin 0.

2(46.43 m/s)sin (45°)
9.8 m/s?
= 6.7 s.

Part a

The time that the arrow spends in the air is 6.7 s.

The time of flight of a projectile is equal to the twice the initial vertical component of the velocity divided by the acceleration due to gravity. The termin the equation1 - 2V sinorepresents the vertical component of the initial velocity.

Step 2 of 2

(b)

The useful kinematic equation of motion that is the relation between distance travelled (s), initial velocity (u), acceleration, and t time is,

Substitute 0 for, g for a, and h for s in the above equation and solve for t.

สี 90 สี 50

Substitute 6 m for h and 9.8 m/sfor g in the above equation.

2(6 m)
V(9.8 m/s)
= 1.11s

The required time is,

t=t-t
= 6.7 s -1.11s
=5.59 s

The time taken to leave the apple from the tree is 5.59 s.

Part b

The time taken to leave the apple from the tree is 5.59 s.

The time taken by the object to reach the ground from the maximum height is known as time of descent. The required time is equal to the time taken to fall the apple minus the time of flight of the arrow.

Answer

Part a

The time that the arrow spends in the air is 6.7 s.

Part b

The time taken to leave the apple from the tree is 5.59 s.

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