question 2
0 pts
a projectile flying through the air has a velocity vector that makes an angle 35.2° with the horizontal at time t = 0s. at the moment it has travelled a horizontal distance of 35.8m, the projectile reaches its peak height. find the time in seconds at which this occurs. provide at least two decimal places

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Accepted Answer

# Explanation:
## Step1: Find the horizontal component of velocity
Let the initial velocity be $v_0$. The horizontal component of velocity $v_{0x}=v_0\cos\theta$, where $\theta = 35.2^{\circ}$.
The horizontal distance $x = v_{0x}t$, so $35.8=v_0\cos(35.2^{\circ})t$.

## Step2: Find the time to reach maximum height
At maximum - height, the vertical component of velocity $v_y = v_0\sin\theta - gt=0$ (where $g = 9.8\ m/s^2$), so $v_0\sin\theta=gt$. Then $v_0=\frac{gt}{\sin\theta}$.

Substitute $v_0$ into the horizontal - distance formula:
$$
\begin{align*}
35.8&=\frac{gt}{\sin\theta}\cos\theta\times t\
35.8&=\frac{9.8t^{2}\cos(35.2^{\circ})}{\sin(35.2^{\circ})}\
35.8&=9.8t^{2}\cot(35.2^{\circ})\
t^{2}&=\frac{35.8}{9.8\times\cot(35.2^{\circ})}\
\cot(35.2^{\circ})&=\frac{\cos(35.2^{\circ})}{\sin(35.2^{\circ})}\approx1.429\
t^{2}&=\frac{35.8}{9.8\times1.429}\
t^{2}&=\frac{35.8}{14.0042}\
t^{2}&\approx2.556\
t&\approx1.60\ s
\end{align*}
$$

# Answer:
$1.60$

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QUESTION IMAGE

Question

Explanation:

Step1: Find the horizontal component of velocity

Step2: Find the time to reach maximum height

Answer: