Question

A long jumper jumps at a 20-degree angle and attains a maximum altitude of 0.6 m. What is her initial speed? [10m/s] How far is her jump? [6.59m]
I need someone to explain to me how to get the answers listed above my showing a detailed step to step process.

Answer 1

Refer to the diagram shown below.

Assume g = 9.8 m/s² and ignore air resistance.

Let V = the initial speed
Let d = the distance raveled (how far she jumps).

The horizontal velocity is
Vx = V cos(20°) = 0.9397V m/s
The vertical launch velocity is
Vy = V sin(20°) = 0.342V m/s

At maximum height of 0.6 m, the vertical velocity is zero. Therefore
Vy² - 2*(9.8 m/s²)*(0.6 m) = 0
(0.342V m/s)² = 11.76 (m/s)²
0.342V = 3.4293
V = 10.027 m/s

Therefore
Vy = 0.342*10.027  = 3.4239 m/s
Vx = 0.9397*10.027 = 9.4226 m/s

The time to attain maximum height is a half of the total time of travel. This time is given by
Vy - gt = 0
3.4239 = 9.8*t
t = 3.4239/9.8 = 0.3494 s
The total travel time is 2*0.3494 = 0.6988 s

The length of the jump is
Vx*(2t) = (9.4226 m/s)*(0.6988 s) = 6.5845 m

Answer:
The initial speed is 10.0 m/s (nearest tenth)
The length of the jump is 6.58 m (nearest hundredth)