1 / 17

170 likes | 281 Views

Aim: How can we approach projectile problems?. Do Now: An object is launched up vertically with an initial velocity of v oy and lands back on the ground in time t. Using the formula, d = v oy t + ½ at 2 , how long will it be in the air for?. Separate Components.

Download Presentation
## Aim: How can we approach projectile problems?

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**Aim: How can we approach projectile problems?**Do Now: An object is launched up vertically with an initial velocity of voy and lands back on the ground in time t. Using the formula, d = voyt + ½ at2, how long will it be in the air for?**Separate Components**• Velocity, acceleration, and distance have both horizontal (x) component, and vertical (y) component**Horizontally Fired Objects**• Horizontal velocity is constant: • There is no horizontal acceleration: a = 0 m/s2 • Horizontal distance (also known as the range) is found using this equation:**Vertical Components**• For the vertical motion, the object is in free fall: vo = 0 m/s • Vertical acceleration = gravity: a = g = 9.8 m/s2 • Vertical distance (also known as the height) is found using the distance equation: y = vot + ½at2**Projectiles Fired at an Angle**• Unlike horizontal projectiles, there are both vox and voy • Initial velocity can be resolved into x and y components vi voy= vosinθ θ vox = vocosθ**Now what?**• Since you can determine your vox and voy, the rest we can figure out!!! • Horizontal velocity is still constant • Horizontal acceleration is still zero • Vertical acceleration is -10 m/s2 (just like in vertically fired objects) • Vertical vf at max height is 0 m/s (just like in vertically fired objects) • Vertical displacement is 0 m (the object starts on the ground and ends on the ground) • Time to max height = ½ total time**1. An object is released from rest on a planet that has no**atmosphere. The object falls freely for 3.0 meters in the first second. What is the magnitude of the acceleration due to gravity on the planet? (A) 1.5 m s2 (B) 3.0 m/s2 (C) 6.0 m/s2 (D) 10.0 m/s2 (E) 12.0 m/s2 No Calculator Allowed **1 minute****2. An object is thrown with a horizontal velocity of 20 m/s**from a cliff that is 125 m above level ground. If air resistance is negligible, the time that it takes for the object to fall to the ground from the cliff is most nearly • 3 s • 5 s • 6 s • 12 s • 25 s No Calculator Allowed **1 minute****3. A 2‑kilogram block rests at the edge of a platform that**is 10 meters above level ground. The block is launched horizontally from the edge of the platform with an initial speed of 3 meters per second. Air resistance is negligible. The time it will take for the block to reach the ground is most nearly • 0.3 s • 1.0 s • 1.4 s • 2.0 s • 3.0 s No Calculator Allowed **1 minute****4. A diver initially moving horizontally with speed v dives**off the edge of a vertical cliff and lands in the water a distance d from the base of the cliff. How far from the base of the cliff would the diver have landed if the diver initially had been moving horizontally with speed 2v? • D • 2d • 4d • can’t be determined without knowing the height of the cliff No Calculator Allowed **1 minute** Velocity is directly proportional to distance 2v = 2d**No Calculator Allowed****1 minute** 5. A projectile is fired with initial velocity vo at an angle 0 with the horizontal and follows the trajectory shown above. Which of the following pairs of graphs best represents the vertical components of the velocity and acceleration v and a, respectively, of the projectile as functions of time t? Vertical acceleration is constant (-10 m/s2)**Questions 6-7**A ball is thrown and follows the parabolic path shown above. Air friction is negligible. Point Q is the highest point on the path. Points P and R are the same height above the ground. 6. How do the speeds of the ball at the three points compare (A) vp < vQ < vR (B) vR < vQ < vP (C) vQ < vR < vP (D) vQ< vP = vR (E) vP = vR < vQ No Calculator Allowed **1 minute** Points P and R have the same height, therefore the same speed (velocity is different) Point Q has less speed (x-comp is the same but there is no y-comp) Resultant is less**7. Which of the following diagrams best shows the direction**of the acceleration of the ball at point P ? (A) (B) (C) (D) (E) Acceleration is due to gravity Gravity always go down No Calculator Allowed **1 minute****A rock of mass m is thrown horizontally off a building from**a height h, as shown above. The speed of the rock as it leaves the thrower’s hand at the edge of the building is v0. 8. How much time does it take the rock to travel from the edge of the building to the ground? (A) (B) (C) (D) (E) Or try dimensional analysis No Calculator Allowed **1 minute****9. A 0.50 kg cart moves on a straight horizontal track. The**graph of velocity v versus time t for the cart is given below. Calculator Allowed **2.5 minutes** From t = 25 s until the cart reaches the end of the track, the cart continues with constant horizontal velocity. The cart leaves the end of the track and hits the floor, which is 0.40 m below the track. Neglecting air resistance, determine each of the following: i. The time from when the cart leaves the track until it first hits the floor ii. The horizontal distance from the end of the track to the point at which the cart first hits the floor

More Related