At a track and field meet, the best long jump is measured as 8.30 m. The jumper took off at an angle of 31 ° to the horizontal. What was the jumper’s initial speed?

let his initial velocity be V.

Then the initiualvertical velocity was V*sin31. Initial horizontal velocity= V*cos31

time in air:
t=8.30m/Vcos31

In the vertical:
hf=hi+vi't-4.9t^2
0=vSin31*8.30m/Vcos31-4.9(8.30m/Vcos31)^2
so, solve for V, a little albebra involved.

or, more directly, recalling that the range is

R = v^2/g sin2θ

just solve for v in

v^2/9.8 * sin62° = 8.3

Well, isn't it impressive that long jumpers can launch themselves into mid-air like human projectiles? Just imagine if we could use that skill to launch ourselves into a nice comfy bed. Ah, the possibilities! Anyway, let's hop into the calculation.

Now, to find the initial speed, we can break it down into horizontal and vertical components. Since the jumper took off at an angle of 31 degrees to the horizontal, we'll need some trigonometry here (don't worry, it won't bite).

The horizontal component of the initial speed remains constant throughout the jump, so it will be the same as the final horizontal speed. Since the jumper travels 8.30 m horizontally, the horizontal component of the initial speed is 8.30 m/s.

Now, let's dive into the math. Using trigonometry's trusty sidekick, the cosine function, we can find the magnitude of the initial velocity.

cos(31 °) = adjacent / hypotenuse

cos(31 °) = 8.30 m/s / hypotenuse

Hypotenuse = 8.30 m/s / cos(31 °)

Hypotenuse ≈ 9.57 m/s

So, the jumper's initial speed is approximately 9.57 m/s. I hope that answer didn't make you jump out of your seat!

To find the jumper's initial speed, we can use the horizontal and vertical components of the motion.

First, let's split the initial velocity into horizontal and vertical components. The horizontal component remains constant throughout the motion and is given by:

Vx = V * cos(θ),

where Vx represents the horizontal component of velocity, V is the initial speed, and θ is the angle of takeoff, which is 31° in this case.

Now, the vertical component of the velocity changes due to gravity. Using the equation of motion:

Vertical Distance = (Vy * t) + (0.5 * g * t^2),

where Vy is the vertical component of velocity, t is the time of flight, and g is the acceleration due to gravity (9.8 m/s^2).

In the vertical direction, at the peak of the jump, Vy = 0. We can solve for t by rearranging the equation:

0 = (Vy * t) + (0.5 * g * t^2).

We can rearrange the equation to solve for t:

0.5 * g * t^2 = -Vy * t,
0.5 * g * t = -Vy,
t = -2Vy / g.

Since the time of flight is equal to twice the time it takes to reach the peak, we can substitute t into our equation and solve for Vy:

t = -2Vy / g,
t = 2 * (V * sin(θ)) / g.

Now we can plug t back into the vertical motion equation:

Vertical Distance = (Vy * t) + (0.5 * g * t^2),
0 = (V * sin(θ)) * 2 * (V * sin(θ)) / g + 0.5 * g * (2 * (V * sin(θ)) / g)^2,
0 = (2 * V^2 * sin^2(θ)) / g - (2 * V^2 * sin^2(θ)) / g + (2 * V^2 * sin^2(θ)) / g,
0 = (2 * V^2 * sin^2(θ)) / g,
0 = (2 * V^2 * sin^2(31°)) / g.

We can solve for V by rearranging the equation:

V = √((g * Vertical Distance) / (2 * sin^2(θ))).

Now we just need to substitute the measured vertical distance, which is the best long jump distance of 8.30 m, the angle of takeoff of 31°, and the acceleration due to gravity of 9.8 m/s^2 into the equation:

V = √((9.8 * 8.30) / (2 * sin^2(31°))).

By evaluating this expression, we find that the jumper's initial speed is approximately 9.598 m/s.