A projectile is launched at ground level with an initial speed of 45.0 m/s at an angle of 30^o (30 degree) above the horizontal. It strikes a target above the ground 2.5 seconds later. What are the and distances from where the projectile was launched to where it lands

1 answer:

Answer:

103.32 m

Explanation:

From the question, The distance from where the projectile was lunched to where its land is the range(R) of the projectile

R = (U²sin2Ф)/g.................... Equation 1

Where U = initial velocity of the projectile, Ф = angle above the horizontal, g = acceleration due to gravity.

Given: U = 45 m/s, Ф = 30°

Constant: g = 9.8 m/s²

Substitute these values into equation 1

R = (45²sin30°)/9.8

R = 2025(0.5)/9.8

R = 103.32 m.

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A.

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