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Free Fall and Projectile Motion Lecture and Sample Problems
- 2. What are theformulas for the following quantities? VELOCITY SPEED ACCELERATION
- 4. DEFINITION • Free fallis the motion of a body when only the force due to gravity is acting on the body. • The acceleration on an object in free fall is called the acceleration due to gravity, or free-fall acceleration. • Free-fall acceleration is the same for all objects, regardless of mass.
- 5. MOREOVER Free-fall acceleration onEarth’s surface is –9.81 m/s2 (-32 ft/s2) at all points in the object’s motion. Consider a ball thrown up into the air. • Moving upward: velocity is decreasing, acceleration is –9.81 m/s2 • Top of path: velocity is zero, acceleration is –9.81 m/s2 • Moving downward: velocity is increasing, acceleration is –9.81 m/s2
- 6.
- 7. What would happento the free fall motion if it were conducted on the Moon, where the acceleration due to gravity is weaker than on Earth?
- 8. 𝑣 = 𝑣0+ 𝑎𝑡 𝑦 = 𝑦0 + 𝑣0𝑡 + 1 2 𝑎𝑡2 𝑣2 = 𝑣0 2 + 2𝑎(𝑦 − 𝑦0) 𝑣 – final velocity (m/s) 𝑣0 – initial velocity (m/s) 𝑎 – acceleration (m/s2) 𝑡 – time (s) 𝑦 – final displacement (m) 𝑦0 – initial displacement (m)
- 9. Equations of Kinematicsat Constant Acceleration
- 10. A piece ofstone was dropped from the top of a tall building. The stone took 10 s to hit the ground. What is the velocity of the stone just before it hit the ground? Given: t = 10 secs. 𝑎 = - 9.81m/s2 𝑣0 = 0 m/s 𝑣 = ? 𝑣 = 𝑣0 + 𝑎𝑡 𝑣 = 𝑣0 + 𝑎𝑡 𝑣 = 0 Τ 𝑚 𝑠 + (−9.81 Τ 𝑚 𝑠 2 )(10𝑠) 𝑣 = (−9.81 Τ 𝑚 𝑠 2 )(10𝑠) 𝑣 = −98.1 𝑚/𝑠 𝑣 = −100 𝑚/𝑠
- 11. A ball isdropped from the top of a tall building. Assuming free fall, how far does the ball fall in 1.50 s. Given: t = 1.50 secs. 𝑎 = - 9.81m/s2 𝑣0 = 0 m/s 𝑦0 = 0 m 𝑦 = ? 𝑦 = 𝑦0 + 𝑣0𝑡 + 1 2 𝑎𝑡2 𝑦 = 𝑦0 + 𝑣0𝑡 + 1 2 𝑎𝑡2 𝑦 = 0 𝑚 + 0 𝑚/𝑠(1.50𝑠) + 1 2 −9.81 m/s2 1.50s 2 𝑦 = 1 2 −9.81 m/s2 1.50s 2 𝑦 = −4.91 m/s2 2.25s2 𝑦 = −11.0475 𝑚 𝑦 = −11.0 𝑚
- 12. A ball isinitially thrown straight up into the air at 50 m/s. (a.) How high does it go? (b.) How long is it in the air? Given: 𝑎 = - 9.81m/s2 𝑣0 = 50 m/s 𝑣 = 0 m/s 𝑦0 = 0 m 𝑦 = ? (a) 𝑡 = ? (b) Solve for (a) 𝑣2 = 𝑣0 2 + 2𝑎(𝑦 − 𝑦0) 0 = 50 Τ 𝑚 𝑠 2 + 2(−9.81 Τ 𝑚 𝑠2)(𝑦 − 0𝑚) 0 = 2500 Τ 𝑚2 𝑠2 + (−19.62 Τ 𝑚 𝑠2)(𝑦) −2500 Τ 𝑚2 𝑠2 = (−19.62 Τ 𝑚 𝑠2)(𝑦) −19.62 𝑚/𝑠2 𝑦 = 127.420999 𝑚 𝑦 = 100 𝑚
- 13. A ball isinitially thrown straight up into the air at 50 m/s. (a.) How high does it go? (b.) How long is it in the air? Given: 𝑎 = - 9.81m/s2 𝑣0 = 50 m/s 𝑣 = 0 m/s 𝑦0 = 0 m 𝑦 = ? (a) 𝑡 = ? (b) Solve for (b) 𝑣 = 𝑣0 + 𝑎𝑡 0 = 50 𝑚/𝑠 + (−9.81 Τ 𝑚 𝑠)(𝑡) −50 𝑚/𝑠 = (−9.81 Τ 𝑚 𝑠)(𝑡) −50 𝑚/𝑠 = (−9.81 Τ 𝑚 𝑠)(𝑡) −9.81 Τ 𝑚 𝑠 𝑡 = 5.096839959𝑠 𝑡 = 5 𝑠