The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7 m when leaving the ground at an angle of 45°. With what speed must the animal leave the ground to reach that height?
Given:
initial velocity =u
angle of projectile=45o
maximum hight achived=3.7m
solution:
according o projectile motion,
maximum hight achived by projectile is:
"h={u^2cos^2\\theta \\over 2g}"
"3.7={u^2 cos^245^o \\over 20}"
"u^2=148"
u=12.6 m/s .............answer
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