Question

Suppose that a bobcat can jump to a height of 1.50 m. It jumps at an angle of 50.00 relative to the horizontal. What speed does the bobcat have to leave the ground in order to reach that height?

Answer 1

Answer: 7.08 m/s

let the bobcat jumps with speed $u_o$ at an angle 50 degrees relative to the horizontal.

Height to be reached, $h=1.50 m$

The component of speed in horizontal direction: $u_o cos 50^o$

The component of speed in vertical direction: $u_o sin 50^o$

We will use equation of motion:

$v^2-u^2=2as$

where v is the final velocity, u is the initial velocity, a is the acceleration, s is the displacement.

The instantaneous speed at the highest point would be 0. Hence, v=0.

The initial velocity in the vertical direction is $u=u_o sin 50^o$

The acceleration would be due to gravity in the downward direction, $a= -g$

The displacement would be the height reached, $s=h=1.50 m$

Insert the values in the equation of motion:

$\Rightarrow 0-(u_o sin50^o)^2=-2gh\\ \Rightarrow u_o^2=\frac{2\times 9.8 m/s^2 \times 1.50 m}{(sin 50^o)^2}=\frac{29.4m^2/s^2}{0.58}=50.1 m^2/s^2\\ \Rightarrow u_o=\sqrt{50.1m^2/s^2}=7.08m/s.$

Hence, the bobcat must leave the ground with the speed of 7.08 m/s at 50 degrees from the horizontal in order to reach the height of 1.50 m .