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?

7.08 m/s

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

Height to be reached,

The component of speed in horizontal direction:

The component of speed in vertical direction:

We will use equation of motion:

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

The acceleration would be due to gravity in the downward direction,

The displacement would be the height reached,

Insert the values in the equation of motion:

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 .

c)as speed changes, wavelength and frequency change.

everything changes!