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!
Dear Rodel,
You're welcome. We are glad to be helpful.
If you liked our service please press like-button beside answer field. Thank you!
Thank you!
Comments
Dear Rodel,
You're welcome. We are glad to be helpful.
If you liked our service please press like-button beside answer field. Thank you!
Thank you!