Assignment # 3 - Acceleration Due to Gravity

Physics 11 Honour Unit 3: Assignment 3.4 Acceleration Due To Gravity

Assignment # 3 - Acceleration Due To Gravity

Name: ______Block:______

Remember to use the value of a = -9.80 m/s2 for these problems.

1.  A rock is dropped from the 100th floor of a tall building (350 m above the ground). How long does it take to fall to the ground below and with what speed will it hit the ground at? 8.45 s; 83 m/s

2.  A ball is thrown straight upwards and reaches a maximum height of 95 m. With what initial velocity was it thrown? 43.2 m/s

3.  A baseball is thrown straight upwards at 28.0 m/s.

a)  To what maximum height will it reach?

40 m

b) How long will it remain in the air (from being thrown upwards to being caught)?

5.7 s

4.  A mischievous person is on an air balloon at a height of 750 m above the ground. He sees his archenemy standing below the balloon and throws a 10.0 kg anvil straight downward with an initial velocity of 15 m/s. With what speed does it hit the unaware enemy? 122 m/s How long it was before the archenemy was hit? 10.9 s

5.  A pebble is thrown vertically downward from a bridge with an initial velocity of 10.0 m/s and strikes the water in 2.0 s. a) Find the velocity with which the pebble strikes 30. m/s and b) the height of the bridge. 40. m

6.  A person throws a ball upward with an initial velocity of 15.0 m/s. Calculate:

a) how high it goes.

11.5 m

b) how long the ball is in the air before it comes back to his hand. 3.06 s

7.  A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth? 1.5 s

8.  A hammer falls out of a helicopter from a height of 3000.0 m. What velocity will the hammer reach just before it hits the ground if it continues to accelerate downwards? 242 m/s How long will it take to hit the ground? 24.7 s

9.  A boy shoots a rock straight up with an initial velocity of 16 m/s. He quickly reloads and shoots another rock in the same way 2.0 s later. Pretty tricky, eh?

a)  At what time and at what height do the rocks meet?

So, the first rock takes 1.63 seconds to reach max height of 13.1 m. That means it has already fallen 0.37 seconds before the second rock is fired. The other rock is following the same pathway (and has same data). Consider the height and velocity of the first rock when the second rock is fired – i.e. after it has fallen for 0.37 seconds. The first rock would be already traveling downward with a velocity of 3.63 m/s and has fallen 0.67 m (now at a height of 12.4 m off of the ground). Here’s the trick …. Given the new set up, the time for the rocks to meet would have to be the same AND the displacements would be related by the difference in height …. So you have to get two equations and solve them as a system … a little mathematics from Math 10! Equations are (using d = vit + 1/2at2):

Rock 1 12.4 - x = 3.63t + 1/2(9.80)t2 watch the signs!!!!

Rock 2 x = 16t + 1/2(-9.80)t2

the x represents the distance that rock 2 will go up and the 12.4 – x represents the distance that rock 1 will fall before they meet.

Put these together and reduce to find the value of t!!

t = 0.63 s after Rock 2 is fired and they meet at a height of 8.1 m

b)  What is the velocity of each rock when they meet? Work from the time

Rock 1 9.8 m/s

Rock 2 9.8 m/s

10.  A water balloon is dropped from the top of a tower, 200.0 m off the ground. An alert archer at the base of the tower sees the balloon and shoots an arrow straight up toward the balloon 5.0 s after the balloon is dropped. The arrow's initial velocity is 40.0 m/s. Where does the arrow intercept the balloon?

Draw a little picture of this.

Think about the water balloon for a moment. Its initial velocity is 0 m/s but after 5.0 seconds it will have fallen 122.5 m (so it is now at a height of 200 - 122.5 = 77.5 m with a velocity of 49 m/s downwards). Then the arrow is fired.

The arrow will go upwards but will travel some distance less than 77.5 m (because the balloon is still falling). But the time for the two objects to reach each other will be the same.

Again you need two equations. One for the balloon dropping and one for the arrow traveling upwards. Plus you need to consider the effect of that 5.0 second time difference:

Equations are (using d = vit + 1/2at2):

Balloon 77.5 – x = 49 t + 1/2(9.80)t2 watch the signs!!!!

Arrow x = 40t + 1/2(-9.80)t2

the x represents the distance that arrow will go up and the 77.5 – x represents the distance that water balloon will fall before they meet.

Put these together and reduce to find the value of t!!

t = 0.87 s (the arrow only reaches a height of 31.1 m before it encounters the balloon)

Are there other ways to do Questions 9 and 10? Probably, I just did it my way!

11.  A ball rolling down an incline travels 6.0 cm in the first 0.25 seconds, and 24 cm in the first 0.50 seconds. Find:

a)  The average speed for the first quarter second time interval

b)  The average speed for the second quarter second time interval.

c) Find its acceleration.

12.  An object has an initial forward velocity of 25 m/s and a constant acceleration of –1.5 m/s2. How far does it travel in 5.0 seconds?

13.  A soccer ball rolls to a stop from a speed of 21 m/s with a deceleration of –2.0m/s2. How far does it roll until it stops?

14.  A skateboarder rolls down Beach Avenue with an initial velocity of 3.5m/s. If her acceleration is 1.5 m/s2, how long will it take her to reach the bottom 120m away?

15.  An object moves for 3.0 seconds with constant acceleration, during which time it travels 81m. The acceleration ceases, and during the next 3.0 seconds it travels 72m.

a)  How would you describe its motion in the last 3.0 seconds?

b)  What was the initial velocity before the acceleration?

16.  A car travelling at 60.0 m/s forward slows down to 40 m/s forward. in 2.5 seconds. Assuming that this acceleration continues, how many seconds would it take the car to slow to

(a) 12 m/s forward?

(b) rest?

17.  A model engine accelerates forward from rest along a straight track at 10.0 m/s2 for 3.0 seconds. It then accelerates forward at 16 m/s2 for 5.0 seconds, and finally decelerates at 12 m/s2 until it stops.

(a)  Calculate how many seconds the engine was moving.

(b)  What was its maximum velocity?

(c) Plot a velocity/time graph of its motion

(d) Find the total distance travelled.

Mr. Marzouk