A projectile is shot horizontally at 25 m/s from the roof of a building 90 m tal

Question

A projectile is shot horizontally at 25 m/s from the roof of a building 90 m tall. Use g = 10 m/s2. Choose positive vertical direction upward. •

How long does it take the projectile to reach the ground below (round to the nearest integer) .

• What is the horizontal range of this projectile, i.e. how far from the base of the building does the projectile land (round to the nearest integer) .

• What is the horizontal component of the projectile’s final velocity with which it hits the ground (round to the nearest integer) . •

What is the vertical component of the projectile’s final velocity with which it hits the ground (round to the nearest integer) .

• What is the magnitude of the projectile’s final velocity (round to the nearest integer) .

• What angle does the final velocity vector make with the horizontal? (round to the nearest integer) degrees.

• What is the horizontal component of the projectile’s acceleration just before it hits the ground (round to the nearest integer) .

• What is the vertical component of the projectile’s acceleration just before it hits the ground (round to the nearest integer) ) .

• What is the magnitude of the projectile’s acceleration just before it hits the ground (round to the nearest integer) .

• What angle does the acceleration vector make with the horizontal? (round to the nearest integer) degrees.

Explanation / Answer

1)

along vertical

initial velocity voy = 0

displacement y = -90

from equation of motion

y = voy*t + (1/2)*ay*t^2

-90 = 0 - (1/2)*10*t^2

t = 4.24 s

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2)
horizontal range x = vox*t = 25*4.24 = 106 m

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3)

vx = vox + ax*t = 25 + 0 = 25 m/s

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4)

vy = voy + ay*t

vy = 0 -10*4.24 = -42.4 m/s

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5)

final velocity =vf = sqrt(vx^2+vy^2) = sqrt(25^2+42.4^2) = 49.22 m/s

====================================

6)

angle = tan^-1(vy/vx) = -59.5

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7)

ax = 0

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8)

ay = -10 m/s^2

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9)

a = sqrt(ax^2+ay^2) = 10 m/s^2

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10)

angle = tan^-1(ay/ax) = -90 degrees

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