A diver runs horizontally off the end of a diving board with an initial speed of 1.95 m/s. If the diving board is 2.00 m above the water, what is the diver's speed just before she enters the water

V = 6.56 m/s

Explanation:

in order to find the speed of the diver before she enters we need to find the vertical component of its final velocity by using 3rd equation of motion:

2gh = Vy² - V₀y²

where,

g = 9.8 m/s²

h = height = 2 m

Vy = Vertical Component of Final Velocity = ?

V₀y = Vertical Component of Initial Velocity = 0 m/s

Therefore,

2(9.8 m/s²)(2 m) = Vy² - (0 m/s)²

Vy = √39.2 m²/s²

Vy = 6.26 m/s

Assuming negligible air friction, the x-component of the final velocity will be the same as the initial velocity:

Vₓ = V₀ₓ = 1.95 m/s

Hence, the final velocity will be:

V = √(Vₓ² + Vy²)

V = √[(1.95 m/s)² + (6.26 m/s)²]

V = 6.56 m/s

f=ma newton

f=100x2.5=250n

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