An arrow is shot with a velocity of 30.0 meters/second and at an angle of 37 degrees above ground. Its initially 2 meters above ground and 15 meters from wall. Wat Height above grond does it hit wall?

How to answer the question

To find the height above the ground at which the arrow hits the wall, we can break down the motion of the arrow into horizontal and vertical components.

First, let's find the time it takes for the arrow to reach the wall. We can use the horizontal component of the velocity (Vx) and the equation of motion:

distance = speed × time

Since the horizontal distance is given as 15 meters and the horizontal component of velocity is given as 30.0 m/s, we have:

15 = 30 × time

Simplifying the equation, we find:

time = 15 / 30
= 0.5 seconds

So, it takes 0.5 seconds for the arrow to reach the wall.

Next, let's find the vertical distance traveled by the arrow during this time. We can use the vertical component of the velocity (Vy) and the equation of motion:

distance = initial velocity × time + (0.5 × acceleration × time^2)

Since the arrow is initially 2 meters above the ground, the initial velocity is the vertical component of the velocity, which can be found using trigonometry:

Vy = velocity × sin(angle)

Plugging in the values, we have:

Vy = 30 × sin(37°)
≈ 18.105 meters/second

Using this value of Vy, we can substitute it into the equation of motion:

distance = (18.105 × 0.5) + (0.5 × 9.81 × 0.5^2)

Simplifying the equation, we find:

distance = 9.0525 + 0.61525
≈ 9.66775 meters

Therefore, the arrow hits the wall at a height of approximately 9.66775 meters above the ground.