A projectile is shot at a velocity of # 16 m/s# and an angle of #pi/6 #. What is the projectile's peak height?

Question

A projectile is shot at a velocity of # 16  m/s# and an angle of #pi/6   #. What is the projectile's peak height?

Nathan Ashcroft · Accepted Answer

The peak height is #=3.27m#

Resolving in the vertical direction #uarr^+#

The initial velocity is #u=16sin(1/6pi)ms^-1#

Let the peak height be #=hm#

At the greatest heignt #v=0ms^-1#

The acceleration due to gravity is #g=9.8ms^-2#

Applying the equation of motion

#v^2=u^2+2as=u^2-2gh#

#h=(u^2-v^2)/(2g)=((16sin(1/6pi))^2-0)/(2g)=3.27m#

Isabella Alvarez · Answer

undefined To find the peak height of the projectile, we can use the following equation:

$$h_{\text{peak}} = \frac{v^2 \sin^2(\theta)}{2g}$$

Where:
- $v$ is the initial velocity of the projectile (16 m/s)
- $\theta$ is the launch angle (π/6 radians or 30 degrees)
- $g$ is the acceleration due to gravity (approximately 9.8 m/s²)

Substituting the given values into the equation:

$$h_{\text{peak}} = \frac{(16 \text{ m/s})^2 \sin^2(\pi/6)}{2 \times 9.8 \text{ m/s}^2}$$

Solving for $h_{\text{peak}}$:

$$h_{\text{peak}} ≈ 7.86 \text{ meters}$$

Therefore, the peak height of the projectile is approximately 7.86 meters.