QUESTION 17 A projectile is shot horizontally at 25 m/s from the roof of a build

Question

QUESTION 17 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) 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 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) What is the horizontal component of the projectile's acceleration just before it hits the ground (round to the nearest integer) degrees 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

velocity, u=25 m/sec

height, h=90 m

a)

use, h=1/2*g*t^2

90=1/2*10*t^2

==> t=4.24 sec

time taken to reach the ground, t=4.24 sec or 4 sec

b)

horizontal range, x=u*t

x=25*4.24

x=381.6 m or 382 m

c)

horizontal component of velocity, ux=25 m/sec^2

d)

vertical component of velocity, uy=g*t

uy=10*4.24

uy=42.4 sec

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.