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(b) We can find the final horizontal and vertical velocities ${v}_{x}$ and ${v}_{y}$ with the use of the result from (a). Then, we can combine them to find the magnitude of the total velocity vector $\overrightarrow{v}$ and the angle $\theta $ it makes with the horizontal. Since ${v}_{x}$ is constant, we can solve for it at any horizontal location. We choose the starting point because we know both the initial velocity and the initial angle. Therefore,
The final vertical velocity is given by [link] :
Since ${v}_{0y}$ was found in part (a) to be 21.2 m/s, we have
The magnitude of the final velocity $\overrightarrow{v}$ is
The direction ${\theta}_{v}$ is found using the inverse tangent:
Of interest are the time of flight, trajectory, and range for a projectile launched on a flat horizontal surface and impacting on the same surface. In this case, kinematic equations give useful expressions for these quantities, which are derived in the following sections.
We can solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface by performing some manipulations of the kinematic equations. We note the position and displacement in y must be zero at launch and at impact on an even surface. Thus, we set the displacement in y equal to zero and find
Factoring, we have
Solving for t gives us
This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. [link] does not apply when the projectile lands at a different elevation than it was launched, as we saw in [link] of the tennis player hitting the ball into the stands. The other solution, t = 0, corresponds to the time at launch. The time of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to g . Thus, on the Moon, where gravity is one-sixth that of Earth, a projectile launched with the same velocity as on Earth would be airborne six times as long.
The trajectory of a projectile can be found by eliminating the time variable t from the kinematic equations for arbitrary t and solving for y ( x ). We take ${x}_{0}={y}_{0}=0$ so the projectile is launched from the origin. The kinematic equation for x gives
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