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A basketball player shoots the ball with a velocity of 17.0 ft/s at an angle of 34.1° with the horizontal. To the nearest tenth, find the magnitude of the horizontal component of the resultant vector.

User TimLer
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We need to find the initial horizontal component of the vector that represents the velocity of the ball. Let us represent that initial horizontal component by,
\bold{v_x}.

Now, since basketball player shoots the ball with an initial velocity,
\bold{v}, of 17.0 ft./sec at an angle of 34.1 degrees with the horizontal, we can find the initial horizontal component by,
\bold{v_x} using the following formula:


\bold{v_x}=\bold{v}\text{cos}(\theta)=17.0*\text{cos}(34.1^\circ)\thickapprox14.1 ft./sec

User ZeusNet
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