A rock is thrown horizontally from the top of a cliff 88 m high, with a horizontal speed of 25 m/s. For what interval of time is the rock in the air
s=ut+12at2s=ut+ \frac{1}{2}at^2s=ut+21at2
88=0t+12×9.8t288=0t+ \frac{1}{2} \times 9.8t^288=0t+21×9.8t2
88=12×9.8t288= \frac{1}{2} \times 9.8t^288=21×9.8t2
88×2×19.8=t288 \times 2 \times \frac{1}{9.8}= t^288×2×9.81=t2
t=88×29.8=4.24st=\sqrt{\frac{88 \times 2}{9.8}} =4.24 st=9.888×2=4.24s
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Nice thanks
Comments
Nice thanks