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2.1 seconds after being projected from ground level, a projectile is displaced 19 m horizontally and 49 m vertically above the launch point. (a) What is the horizontal component of the initial velocity of the particle? 9.04 Correct: Your answer is correct. m/s (b) What is the vertical component of the initial velocity of the particle?

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jolis1796

Answer:

a) v₀x = 9.05 m/s  (→)

b) v₀y = 33.63 m/s  (↑)

Explanation:

Given

t = 2.1 s

x = 19 m

y = 49 m

a) v₀x = ?

We can apply

x = vx*t = v₀x*t  ⇒   v₀x = x/t = (19 m) / (2.1 s) = 9.05 m/s

⇒   v₀x = 9.05 m/s  (→)

b) v₀y = ?

Let's use this equation

y = y₀ + v₀y*t - 0.5*g*t²

⇒  49 m = 0 m + v₀y*(2.1 s) - 0.5*(9.81 m/s²)*(2.1 s)²

⇒  v₀y = 33.63 m/s  (↑)

If we get ymax as follows

ymax = v₀y² / (2*g) = (33.63 m/s)² / (2*9.81 m/s²) = 57.66 m

we can see that  57.66 m > 49 m  (ymax > y)  

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