Question

a projectile flying through the air has a velocity vector that makes an angle 50.3° with the horizontal at time t = 0s. at the moment it has travelled a horizontal distance of 25.5m, the projectile reaches its peak height. find the time in seconds at which this occurs. provide at least two decimal places

Explanation:

Step1: Recall peak time property

At peak height, vertical velocity $v_y = 0$. The time to peak is $t = \frac{v_0\sin\theta}{g}$, where $v_0$ is initial speed, $\theta=50.3^\circ$, $g=9.81\mathrm{m/s^2}$.

Step2: Horizontal distance relation

Horizontal distance $x = v_0\cos\theta \cdot t$. Substitute $v_0\sin\theta = gt$ into this:
$x = \frac{gt}{\tan\theta} \cdot t = \frac{gt^2}{\tan\theta}$

Step3: Solve for $t$

Rearrange to solve for $t$:
$t = \sqrt{\frac{x\tan\theta}{g}}$
Substitute $x=25.5\mathrm{m}$, $\theta=50.3^\circ$, $g=9.81\mathrm{m/s^2}$:
$\tan(50.3^\circ) \approx 1.203$
$t = \sqrt{\frac{25.5 \times 1.203}{9.81}} = \sqrt{\frac{30.6765}{9.81}} = \sqrt{3.127} \approx 1.77$

Answer:

1.77