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Vinil7 [7]
3 years ago
13

Please help on this one

Physics
1 answer:
Sphinxa [80]3 years ago
6 0

Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.

Let's break them down into components.

                X                                Y

v₁     32 cos50 m/s           32 sin50 m/s

v₂     32 cos50 m/s                    ?

Δd             ?                                0

Δt              ?                                ?

a                0                         -9.8 m/s²


Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.

                           Δdy = v₁yΔt + 0.5ay(Δt)²

                                0 = v₁yΔt + 0.5ay(Δt)²

                                0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0

                                0 = v₁ + 0.5ayΔt

                                0 = 32sin50m/s + 0.5(-9.8m/s²)Δt

                                0 = 2<u>4</u>.513 m/s - 4.9m/s²Δt

                 -2<u>4</u>.513m/s = -4.9m/s²Δt

-2<u>4</u>.513m/s  ÷ 4.9m/s² = Δt

                          <u>5</u>.00s = Δt


Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.

Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²

Δdₓ = 32cos50m/s(<u>5</u>.00s) + 0.5(0)(<u>5</u>.00)²

Δdₓ = 32cos50m/s(<u>5</u>.00s)

Δdₓ = 10<u>2</u>.846


Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.






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