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A projectile is launched at ground level with an initial speed of 53.0 m/s at an angle of 35.0° above the horizontal. It strikes a target above the ground 2.50 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

User YoBre
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1 Answer

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Answer:

x = 108.5 m y = 45.37 m,

Step-by-step explanation:

This is a projectile launching exercise, the horizontal distance it reaches is called the range and can be found by the expression

x = v₀ₓ t

the initial hairiness is

v₀ₓ = v₀ cos θ

x = v₀ cos θ t

let's calculate

x = 53 cos 35 2.50

x = 108.5 m

the height of the projectile can be calculated with

y =v₀ t - ½ g t²

y = vo sin tea t - ½ g t²

y = 53 sin 35 2.50 - ½ 9.8 2.5²

y = 45.37 m

User Gaurav Pandit
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