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.
the horizontal component of its velocity vx=vcosθ=10×cos30=8.66msecv _x = vcosθ = 10 \times cos 30 = 8.66 \frac{m}{sec}vx=vcosθ=10×cos30=8.66secm
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}v=vx+at=8.66+1×0.5=9.16secm
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!