Question

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how do height and velocity change the range and time of flight for a horizontal projectile?
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determine how changes in initial conditions affect the trajectory of a projectile
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given d_y = 30m for a horizontal projectile. sketch the motion of the projectile including the horizontal and vertical components of its velocity find the initial velocity in the y direction and the time it is in the air.

Explanation:

Step1: Determine initial vertical velocity

For a horizontal - launched projectile, the initial vertical velocity $v_{0y}=0\ m/s$.

Step2: Use vertical - motion equation to find time of flight

The vertical displacement equation is $d_y = v_{0y}t+\frac{1}{2}gt^{2}$. Since $v_{0y} = 0\ m/s$, the equation simplifies to $d_y=\frac{1}{2}gt^{2}$. We know that $d_y = 30m$ and $g = 9.8m/s^{2}$. Rearranging for $t$ gives $t=\sqrt{\frac{2d_y}{g}}$.
Substitute $d_y = 30m$ and $g = 9.8m/s^{2}$ into the formula: $t=\sqrt{\frac{2\times30}{9.8}}\approx\sqrt{6.12}\approx2.47s$.

Answer:

Initial vertical velocity $v_{0y}=0\ m/s$, time of flight $t\approx2.47s$