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 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 .

Related Questions


Catherine likes the name she was given, but wishes it were shorter. She decides to start going by the new name of Cathy. It's cute, short, easier to say, and more like the names of her friends. Cathy feels like her name fits in now. Which term describes why Catherine shortened her name?



Social conformity


Social conformity is when a person beliefs o acts in a certain way to fit in a group. This is when a person agrees with everyone else because he/she wants to be like the others to feel accepted. According to this, the term that describes why Catherine shortened her name is social conformity because despite that she likes her name, she decided to change it to make it similar to the names of her friends and like that, feel that she fits in.


A car is travelling at a speed of 21m/s. It accelerates at an average rate of 3m/s^2 for a time of 4 seconds. Find the distance it travels.​



S = 108m


According to Second Law of Motion:

S = Vi*t + 0.5*a*t^2

Vi = 21m/s

a = 3m/s^2

t = 4s

Putting all values in Equation

S = 21*4 + 0.5*3*4^2

S = 84 + 24

S = 108m


An iceboat is at rest on a frictionless frozen lake when a sudden wind exerts a constant force of 200. N, towards the east, on the boat. Due to the angle of the sail, the wind causes the boat to slide in a straight line for a distance of 8.0 m in a direction 20 degrees north of east. What is the kinetic energy of the iceboat at the end of that 8.0 m?


K.E = 1600 J


The force exerted on boat is 200 N , which gives the acceleration to it = 200/m

here m is the mass of the boat .

The velocity after covering a distance of 8 m with this acceleration van be found by the relation

v² - u² = 2 a s

here u is the initial velocity , which is zero in this case because boat is at rest .

The v² = 2 x 200/m x 8

The K.E = 1/2 m v² = 1/2 m x 2 x 200/m x 8 = 1600 J


1,600J IS INCORRECTThe kinetic energy is 1503.503j. --> 1,500j

If you looked at this and thought, "bro what the heck 1/2mv^2 doesnt work",

you're right. The kinetic energy formula doesn't work directly.

However, remember that the change in kinetic energy is equal to work. If our initial spot is at rest, that means we only care about the final.

Here's the equation for change in KE: (1/2 m*vf^2) - (1/2 m*vi^2)

Ignore initial because the boat is at rest -- vi = 0

change in KE is now : (1/2 m*vf^2)

Now remember: change in KE = Work

Work = F*d*cos(theta)

Plug in 20 degrees for theta, plug in 200N for force, and 8m for D.

Work = cos(20) * 200N * 8m

Work = 1503.508 = KE

Work ~ 1500 = KE

KE ~ 1500


A spring is compressed between two cars on a frictionless airtrack. Car A has four times the mass of car B, MA = 4 MB, while the spring’s mass is negligible. Both cars are initially at rest. When the spring is released, it pushes them away from each other. Which of the following statements correctly describes the velocities, the momenta, and the kinetic energies of the two cars after the spring is released? Note: Velocities and momenta are given below as vectors. 1. ~vA = −~vB ~pA = −4~pB KA = 16 KB 2. ~vA = + 1 5 ~vB ~pA = + 4 5 ~pB KA = 4 25 KB 3. ~vA = + 1 4 ~vB ~pA = +~pB KA = 4 KB 4. ~vA = − 1 4 ~vB ~pA = −~pB KA = 1 4 KB 5. ~vA = −2~vB ~pA = −8~pB KA = 16 KB 6. ~vA = −~vB ~pA = −~pB KA = KB 7. ~vA = − 1 3 ~vB ~pA = − 2 3 ~pB KA = 4 3 KB 8. ~vA = − 1 2 ~vB ~pA = −2~pB KA = KB 9. ~vA = −4~vB ~pA = −16~pB KA = 64 KB 10. ~vA = − 1 4 ~vB ~pA = −~pB KA = 4 KB



10. True.     vA = - ¼ vB ,  pA = - pB


Let us propose the solution of this problem, as the cars are released let us use the conservation of the moment.

Initial before releasing the cars

         p₀ = 0

Final after releasing cars

        = mA vA + mB vB

        p₀ =

        0 = mA vA + mB vB

        vA = - mB / mA vB


They indicate that mA = 4 mB

        vA = - ¼ vB

Let's write the amount of movement for each body

        pA = mA vA = 4 mB (- ¼ vB

        pA = -mB vB

        pB = mB vB

        pA = - pB

Let's check the answers

1 False

2 False

3 False

4 false

5 False

6 False

7 False

8 False

9 False

10. True. The speed and amount of movement values ​​are correct

Random Questions