Question

1. -/1 points a brick is thrown upward from the top of a building at an angle of 35° to the horizontal and with an initial speed of 11 m/s. if the brick is in flight for 3.4 s, how tall is the building? details my notes ask your teacher

Explanation:

Step1: Resolve the initial velocity into vertical - component

The initial vertical velocity $v_{0y}=v_0\sin\theta$, where $v_0 = 11\ m/s$ and $\theta = 35^{\circ}$. So $v_{0y}=11\sin35^{\circ}\approx11\times0.574 = 6.314\ m/s$.

Step2: Use the kinematic equation for vertical displacement

The kinematic equation for vertical displacement $y = v_{0y}t-\frac{1}{2}gt^{2}$, where $t = 3.4\ s$ and $g = 9.8\ m/s^{2}$.
Substitute the values: $y=(6.314\times3.4)-\frac{1}{2}\times9.8\times(3.4)^{2}$.
First, calculate $6.314\times3.4 = 21.4676$.
Second, calculate $\frac{1}{2}\times9.8\times(3.4)^{2}=4.9\times11.56 = 56.644$.
Then $y=21.4676 - 56.644=- 35.1764\ m$. The negative sign just means the final position is below the starting - point. The height of the building $h=\vert y\vert\approx35.2\ m$.

Answer:

$35.2\ m$