If a projectile is shot at an angle of #(5pi)/12# and at a velocity of #16 m/s#, when will it reach its maximum height??

Question

If a projectile is shot at an angle of #(5pi)/12# and at a velocity of #16   m/s#, when will it reach its maximum height??

Layla Ahmed · Accepted Answer

1.545 seconds

#T = (u sin θ) / g# #T = (16 ms^-1 sin ((5 π)/12)) / (10 ms^-2)# #T = (16 × 0.9659) / (10)# s #T = 1.545# #text(seconds)#

Nathan Ashcroft · Answer

undefined The projectile will reach its maximum height at the time when its vertical velocity component becomes zero. The time taken to reach maximum height can be calculated using the formula: $ t = \frac{v \cdot \sin(\theta)}{g} $, where $ v $ is the initial velocity (16 m/s), $ \theta $ is the launch angle $(5\pi/12)$, and $ g $ is the acceleration due to gravity (9.8 m/s²).

Plugging in the values:
$ t = \frac{16 \cdot \sin(5\pi/12)}{9.8} $

$ t ≈ 1.03 $ seconds.

So, the projectile will reach its maximum height approximately 1.03 seconds after being launched.

Carson Alexander · Answer

undefined To find the time when the projectile reaches its maximum height, you can use the formula for the time taken to reach maximum height for a projectile launched at an angle:

$$ t_{\text{max}} = \frac{V_0 \sin(\theta)}{g} $$

Where:
- $ V_0 $ is the initial velocity (16 m/s in this case)
- $ \theta $ is the launch angle ($ \frac{5\pi}{12} $ in this case)
- $ g $ is the acceleration due to gravity (approximately $ 9.8 \, \text{m/s}^2 $ on the surface of the Earth)

Plug in the given values and calculate to find $ t_{\text{max}} $.