Question

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0 above the horizontal, some of the tiny critters have reached a maximum height of 52.7 above the level ground.
A)What was the takeoff speed for such a leap? v= 3.79m/s
B) What horizontal distance did the froghopper cover for this world-record leap?

Answer 1

Answer:

A) 3.79 m/s  B) 1.33 m

Explanation:

A)

• Horizontal movement:
• Once in the air, no forces act on the froghopper, so it keeps moving with the same initial horizontal speed.
• This horizontal component, is the projection of the velocity vector on the horizontal direction (x-axis):

        $v_{ox} = v_{o} *cos (58.0 deg)$

• The horizontal displacement can be simply calculated as follows:

        $x = v_{ox} *t$

• Vertical movement:
• As the vertical and horizontal are independent each other (due to they are perpendicular, so there is no projection of one movement on the other), in the vertical direction, all happens as if would be a body thrown upward with a given initial vertical velocity.
• This velocity can be found as the projection of the velocity vector on the vertical direction (y-axis):

        $v_{oy} = v_{o} *sin (58.0 deg) (1)$

• Once in the air, the gravity will cause that the froghopper be slow down, till it reaches to the maximum height, where it will come momentarily to an stop.
• In that moment, we can apply the following kinematic equation:

        $v_{fy} ^{2} -v_{oy} ^{2} = 2*g*h_{max}$

• where vfy = 0, g = -9.8m/s2, hmax = 52.7 cm= 0.527 m
• Replacing by the givens, we can solve for voy:

        $v_{oy} =\sqrt{2*g*h_{max}} = \sqrt{2*9.8m/s2*0.527m} =3.21 m/s$

• From the equation (1), we can solve for the magnitude of the initial velocity, v₀:

        $v_{o} = \frac{v_{oy}}{sin 58.0} =\frac{3.21m/s}{0.848} = 3.79 m/s$

B)

• With the value of the magnitude of the initial velocity, we can find the horizontal component vox, as follows:

        $v_{ox} = v_{o} *cos (58.0 deg) =\\ \\ 3.79 m/s * cos (58.0deg) = 2.01 m/s$

• In order to know the horizontal distance travelled, we need to find the time that the insect was in the air.
• We can use the equation for the vertical displacement, replacing this value by 0, as follows:

       $y = 0 = v_{oy} *t -\frac{1}{2} * g *t^{2}$

• Replacing by  the givens, and rearranging terms, we can solve for t:

        $t_{air} =\frac{2*v_{oy} }{g} = \frac{2*3.21 m/s}{9.8 m/s} = 0.66 s$

• Finally, we find the horizontal displacement, as follows:

       $x_{max} = v_{ox} *t_{air} = 2.01 m/s * 0.66 s \\ \\ x_{max} = 1.33 m$

• The horizontal distance covered by  the froghopper was 1.33 m.