Question

A squirrel jumps into the air with a velocity of 7 m/s at an angle of 20 degrees. What os the maximum height reached by the squirrel.

Answer 1

$Given:\\v_0=7\frac{m}{s}\\\alpha =20^\circ\\g=10\frac{m}{s^2} \\\Find:\\H=?\\\\Solution:\\\\ H=\frac{v^2_0\sin^2\alpha}{2g} \\\\H=\frac{(7\frac{m}{s})^2\cdot (\sin(20^\circ))^2}{2\cdot 10\frac{m}{s^2}} \approx 0.3m\\\\\\Proof\;that\;H=\frac{v^2_0\sin^2\alpha}{2g}\;in \;appendage$

Answer 2

Answer:

0.3 m

Explanation:

I used the formula for range which is: $R = \frac{v^2}{g} * sin(2*theta)$

v = 7, theta = 20, g = 9.8 so I just plugged it in and solved