Find Velocity of a long jumper

[dmkeddy]

Homework Statement

An athlete executing a long jump leaves the ground at an angle of 30 degrees and travels 8.90m. What was the take-off speed?

Homework Equations

d(vertical)=(vi)(t) + 0.5at^2

The Attempt at a Solution

0=Vsin30t + (0.5)(-9.8)(t^2)

I cannot solve for two separate variables, how can I solve this problem?

 

[JaredJames]

EDIT: I see what you've done.

Right, in your attempt, d does not equal 0. There has to be some vertical height.

For horizontal you know:

d = 8.9m

For vertical you know:

a = -9.8 for the ascent and 9.8 for the descent.

You know vf for the ascent = 0 and vi for the descent = 0.

 

[dmkeddy]

that is the vertical, there is no horizontal acceleration
Vertical:
Vi=?
Vf=?
a=-9.8m/s^2
t=?
d=0

Horizontal:
Vave=?
t=?
d=8.90m

 

[JaredJames]

that is the vertical, there is no horizontal acceleration
Vertical:

d=0

d does not equal 0. There has to be vertical height.

The vertical phase is split in two. See previous post for details.

Unless you know the flight time, this question all comes down to the vertical components.

 

[rwisz]

To find the take-off speed of the long jumper, we need to use the equation for horizontal distance traveled, which is given by:

d(horizontal) = v(cosθ)t

Where v is the initial velocity, θ is the angle at which the athlete leaves the ground, and t is the time in flight.

In this case, we know the horizontal distance traveled (8.90m), the angle (30 degrees), and we need to solve for v.

Substituting in the values, we get:

8.90m = v(cos30)(t)

We also know that the time in flight is the same for both the horizontal and vertical components, so we can use t from the vertical equation (t = 2.25s) and substitute it into the horizontal equation:

8.90m = v(cos30)(2.25s)

Solving for v, we get:

v = 8.90m / (cos30)(2.25s)

v = 7.87 m/s

Therefore, the take-off speed of the long jumper was 7.87 m/s.