Question #206818

A ball is thrown at angle of 30° above the horizontal with a speed of 10 m/s. After 0.5 sec, find the horizontal component of its velocity.


Expert's answer

the horizontal component of its velocity vx=vcosθ=10×cos30=8.66msecv _x = vcosθ = 10 \times cos 30 = 8.66 \frac{m}{sec}

after 5 sec

Velocity v=vx+at=8.66+1×0.5=9.16msecv = v_x + at = 8.66 + 1\times 0.5 = 9.16 \frac{m}{sec}


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