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A lovesick lad wants to throw a bag of candy and love notes into the open window of his girlfriend’s bedroom 10.0 m above. Assuming it just reaches the window, he throws the love gift at 60.0o to the ground. How far from the house is he standing when he throws the bag?

Respuesta :

hamzaahmeds

Answer:

D = 0.5 m

Explanation:

This motion can be represented as a projectile motion with the following parameters;

θ₀ = Launch Angle = 60°

H = Maximum Height = 10 m

We will use the formula for maximum height of the projectile to get the launch speed (V₀) of projectile:

H = V₀² Sin²θ/2g

Therefore,

10 m = V₉² Sin²60°/(2)(9.8 m/s²)

(10 m )(9.8 m/s²)/(Sin² 60°) = V₀²

V₀ = √130.67 m²/s²

V₀ = 11.43 m/s

Now, we calculate the distance of lad from house (D) that will be equal to half of range of projectile (R):

D = R/2 = V₀² Sin 2θ/2g

D = (11.43 m/s)² Sin 120°/(2)(9.8 m/s²)

D = 0.5 m

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