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Tresset [83]
3 years ago
5

A projectile is fired from the ground at a velocity of 30.0 m/s, 35.0 º from the horizontal. What is the maximum height the proj

ectile reaches in the air? How far away from the point it was fired does it hit the ground? What is the time the projectile is in the air?
Physics
1 answer:
Norma-Jean [14]3 years ago
8 0

Answer:

Vy = V sin theta = 30 * ,574 = 17.2 m/s

t1 = 17.2 / 9.8 = 1.76 sec to reach max height

Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m

H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m

Time to fall from zero speed to ground = rise time = 1.76 sec

Vx = V cos 35 = 24.6 m / sec     horizontal speed

Time in air = 1.76 * 2 = 3.52 sec before returning to ground

S = 24.6 * 3.52 = 86.6 m

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Answer:

2.4 m

Explanation:

Consider the motion along the vertical direction

y_{o} = initial position of ball above the ground = 4.5 m

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y = position of ball at the time of hitting the smokestack

Using the kinematics equation

y = y_{o} + v_{oy} t + (0.5) a_{y} t^{2}

inserting the above values

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We are given that

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