Question

PROBLEM A long jumper (see figure) leaves the ground at an angle of 20.0°to the horizontal and at a speed of 11.0 m/s. (a) How long does it take for him to reach maximum height? (b) What is the maximum height? (c) How far does he jump? (Assume his motion is equivalent to that of a particle, disregarding the motion of his arms and legs.) (d) Use the proper equation to find the maximum height he reaches.

STRATEGY Again, we take the projectile equations, fill in the known quantities, and solve for the unknowns. At the maximum height, the velocity in the y-direction is zero, so setting the proper equation equal to zero and solving gives the time it takes him to reach his maximum height. By symmetry, given that his trajectory starts and ends at the same height, doubling this time gives the total time of the jump.

SOLUTION

(A) Find the time t_max taken to reach maximum height.

Set v_y = 0 in the equation and solve for t_max.

v_y = v₀ sin θ₀ − gt_max = 0

(1)

tmax	=	v0 sin θ0 g
	=	(11.0 m/s)(sin 20.0°) 9.80 m/s2 = 0.384 s

(B) Find the maximum height he reaches.

Substitute the time t_max into the equation for the y-displacement.

y_max = (v₀ sin θ₀)t_max −

1

2

g(t_max)²

ymax	=	(11.0 m/s)(sin 20.0°)(0.384 s)
		− 1 2 (9.80 m/s2)(0.384 s)2

y_max =

0.722 m (C) Find the horizontal distance he jumps.

First find the time for the jump, which is twice t_max.

t = 2t_max = 2(0.384 s) = 0.768 s

Substitute this result into the equation for the x-displacement.

(2)

Δx	=	(v0 cos θ0)t
	=	(11.0 m/s)(cos 20.0°)(0.768 s)
	=	7.94 m

(D) Use an alternate method to find the maximum height.

Use the equation to the right, solving for Δy.

vy2 − v0y2	=	−2gΔy
Δy	=	vy2 − v0y2 −2g

Substitute v_y = 0 at maximum height, and the fact that v_0y = (11.0 m/s) sin 20.0°.

Δy =

0 − [(11.0 m/s) sin 20.0°]²

−2(9.80 m/s²)

= 0.722 m

LEARN MORE

REMARKS Although modeling the long jumper's motion as that of a projectile is an oversimplification, the values obtained are reasonable.

QUESTION How would the time of the jump and the horizontal distance traveled change if g were changed, for example if the jump could be repeated with the same initial velocity on a different planet? (Select all that apply.)

_Increasing the time of the jump has no effect on the displacement.

_The displacement increases with increased time of the jump.

_The displacement decreases with increased time of the jump.

_The time of the jump decreases when g is smaller.

_The time of the jump increases when g is smaller.

PRACTICE IT

Use the worked example above to help you solve this problem. A long jumper (as shown in the figure) leaves the ground at an angle of 17.0° to the horizontal and at a speed of 9.8 m/s.

(a) How long does it take for him to reach maximum height?
_____s

(b) What is the maximum height?
_____ m

(c) How far does he jump? (Assume that his motion is equivalent to that of a particle, disregarding the motion of his arms and legs.)
_____ m

EXERCISE HINTS:  GETTING STARTED  |  I'M STUCK!

A grasshopper jumps a horizontal distance of 1.10 m from rest, with an initial velocity at a 46.0° angle with respect to the horizontal.

(a) Find the initial speed of the grasshopper.
_____m/s

(b) Find the maximum height reached.
_____ m

Answer 1

using equation S = u t + 1/2 g t²

but u = 0 then

S = 1/2 g t²

so using this equation

_Increasing the time of the jump has no effect on the displacement. - False

_The displacement increases with increased time of the jump. - True

_The displacement decreases with increased time of the jump. - False

_The time of the jump decreases when g is smaller. - False

_The time of the jump increases when g is smaller. -True