How to find max height projectile motion
The Correct Answer and Explanation is:
Correct Answer:
To find the maximum height of a projectile in motion, use the following formula:H=v2sin2(θ)2gH = \frac{{v^2 \sin^2(\theta)}}{{2g}}H=2gv2sin2(θ)
Where:
- HHH is the maximum height
- vvv is the initial velocity
- θ\thetaθ is the angle of projection
- ggg is the acceleration due to gravity (approximately 9.8 m/s29.8 \, \text{m/s}^29.8m/s2)
- sin\sinsin is the sine trigonometric function
Explanation
Projectile motion refers to the motion of an object thrown into the air that moves under the influence of gravity. One important characteristic of this motion is the maximum height the object reaches before it starts falling back to the ground.
To calculate this maximum height, we analyze only the vertical component of the initial velocity. The initial velocity vvv is split into two components:
- Horizontal component: vx=vcos(θ)v_x = v \cos(\theta)vx=vcos(θ)
- Vertical component: vy=vsin(θ)v_y = v \sin(\theta)vy=vsin(θ)
At the peak of the motion, the vertical velocity becomes zero because gravity slows the object down until it momentarily stops moving upward before descending.
Using one of the kinematic equations, we can find the height at which this happens:vy2=vy02−2gHv_y^2 = v_{y0}^2 – 2gHvy2=vy02−2gH
Since the final vertical velocity vyv_yvy at the highest point is zero, and vy0=vsin(θ)v_{y0} = v \sin(\theta)vy0=vsin(θ), the equation becomes:0=(vsin(θ))2−2gH0 = (v \sin(\theta))^2 – 2gH0=(vsin(θ))2−2gH
Solving for HHH, we get:H=(vsin(θ))22gH = \frac{{(v \sin(\theta))^2}}{2g}H=2g(vsin(θ))2
This gives us the maximum vertical distance the projectile reaches above its original position.
Note that the angle θ\thetaθ is in degrees or radians, and must be consistent with your calculator’s settings. This formula assumes the projectile is launched from and lands at the same height. If launched from a different height, further adjustments must be made.
