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

0.3 m

Step-by-step explanation:

I used the formula for range which is:
`R = (v^2)/(g) * sin(2*theta)`

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

Answer 2

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