Question

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.
1) Find ta, the time that the arrow spends in the air.
Answer numerically in seconds, to two significant figures.

2) Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree.

Answer 1

Let us solve the Part A first. 

In order to find the time we use the formula for the range of a projectile:

R = v^2 multiply by sin(2θ) divided by g

220 = v^2 multiply by sin(2 multiply by 45) divided by 9.8

220 = v^2 divided by 9.8

V^2 = 2,156

v = 46.43 m/s

Now we will find the x component of the velocity:

vx = sin(θ) multiply by v

vx = cos(45) multiply by 46.43

vx = 32.83 m/s

So the time it takes for the arrow to travel 220 m is the same as the time it is in the air:

time = d divided by  v

time = 220 divided by 32.83

time = 6.70 seconds

6.70 s.

Now we will solve the part B.

We will start by finding the time it takes for the apple to fall 6.0 m

d = v0 multiply by t + 1 divided by 2 at 2 square

6 = 0 multiply by t + 1 divided by 2 multiply by 9.8 multiply by t^2

6 = 4.9 multiply by t^2

t^2 = 1.2245

t = 1.1066 s

Thus the arrow needs 6.70 s to reach the tree, so we will just subtract for the difference:

wait = 6.70 – 1.1066

wait = 5.59 s

Hence the apple should be dropped 5.59 s after the arrow is shot.

I hope it helped

Answer 2

According to the given data, the time that the arrow spends in the air is 6.70 seconds, and the time someone has to drop the apple after the arrow is shot, so that the arrow hits the apple, is 5.60 seconds.

Further explanation

Part 1. First let's focus on the first question. This is a 2D problem (since the arrow moves in 2 directions), therefor we can write equations for both dimensions in which the arrow moves. Let's call x the distance travelled by the arrow on the horizontal direction and y the distance travelled by the arrow on the vertical direction.

Lets suppose that the arrow was shot from a point which we will call the origin (meaning zero horizontal and vertical displacement), and let's call ta the time spent by the arrow on the air. The equations of motion for the arrow will be:

$x= V \cdot cos(45 \º) \cdot t$

$y= V \cdot sin(45 \º) \cdot t- \frac{g \cdot t^2}{2}$

Where V is the initial speed of the arrow. If we evaluate both of the above expressions at ta, we know that its horizontal displacement is 220 m, and its vertical displacement is 0 m. Therefor:

$220= V \cdot cos(45 \º) \cdot ta$

$0= V \cdot sin(45 \º) \cdot ta- \frac{g \cdot {ta}^2}{2}$

Therefor we have a system of 2 unknowns (V and ta) and 2 equations. Solving for V from the first equation and substituting on the second we find that:

$0= 220 \cdot tan(45 \º) - \frac{g \cdot {ta}^2}{2}$

Solving for ta we find that:

$ta= \sqrt{\frac{2 \cdot 220 \cdot tan(45 \º)}{g}}$

Plugging in numerical values, we get that ta is 6.70 seconds.

Part 2. To know the exact moment at which we need to drop the apple so that the arrow hits it, we need to compute that time it takes to fall those 6 meters. Since the apple is dropped with zero initial speed, we can find the time it takes to fall as:

$t= \sqrt{\frac{2 \cdot 6}{g}}$

Which is around 1.10 seconds. Therefor we need to drop the apple 5.60 seconds after the arrow is shot.

Learn more

• Concepts of projectile movement: brainly.com/question/2818076
• Concept of free fall: brainly.com/question/1708231

Keywords

Free falling objects, projectile, gravity