A long jumper leaves the ground at an angle of 20 degrees and travels a horizontal distanceof 7.0m before landing. What is the takeoff speed of the jumper?

To be honest I really don't know how to solve it. A step-by-step explantion would be helpful.

Horizontal:

7=viCos20*timeinair
timeinair=7/Vicos20

Vertical:
hf=hi+ViSin20*t-4.9t^2
put the timeinair expression in for time
0=0+ViSin20(7/cos20)-4.9(49/cos^2 20)

compute Vi

To solve this problem, we can use the principles of projectile motion. Let's break it down step by step:

Step 1: Understand the problem
The problem asks for the takeoff speed of the jumper. We are given the angle of takeoff (20 degrees) and the horizontal distance traveled (7.0m). We need to find the initial speed of the jumper.

Step 2: Identify the relevant equations
Since we are dealing with projectile motion, we can use the following kinematic equations:

- Horizontal velocity (Vx) = initial speed (V0) * cos(angle)
- Vertical velocity (Vy) = initial speed (V0) * sin(angle)

Step 3: Apply the equations to find the initial speed
We know that the horizontal distance traveled (7.0m) is equal to the horizontal velocity multiplied by the time of flight. Since there is no acceleration in the horizontal direction, we can write:

7.0m = V0 * cos(20 degrees) * t

Similarly, we can find the time of flight using the vertical motion equation:

0 = V0 * sin(20 degrees) * t - (g * t^2) / 2

Here, 'g' represents the acceleration due to gravity (approximately 9.8 m/s^2).

Step 4: Solve the equations
We have two equations and two unknowns (V0 and t). We can solve them simultaneously to find the values.

From the first equation, we can isolate t:

t = 7.0m / (V0 * cos(20 degrees))

Substituting this value of t in the second equation:

0 = V0 * sin(20 degrees) * (7.0m / (V0 * cos(20 degrees))) - (g * (7.0m / (V0 * cos(20 degrees))))^2 / 2

This equation can be rearranged to isolate V0:

0 = V0 * sin(20 degrees) * (1 / cos(20 degrees)) - (g * (7.0m)^2) / (V0^2 * (cos(20 degrees))^3) / 2

Simplifying:

V0 = (g * (7.0m)^2 * (cos(20 degrees))^3) / (2 * sin(20 degrees) * cos(20 degrees))

Step 5: Calculate the result
Using the given values of 'g', 'm', and the trigonometric functions, we can plug them into the equation to find the value of V0:

V0 = (9.8 m/s^2 * (7.0m)^2 * (cos(20 degrees))^3) / (2 * sin(20 degrees) * cos(20 degrees))

Using a calculator or math software, you can evaluate this expression to find the takeoff speed of the jumper.