Question

# Basic Physics: Projectile Motion. 1. Given that a ball is launched perfectly horizontally from a height...

Basic Physics: Projectile Motion.

1. Given that a ball is launched perfectly horizontally from a height of 1.2 meters, and lands on the floor a horizontal distance of 2.6 meters away, find the initial speed (or "muzzle velocity") of the ball in meters per second. Use 9.82 meters per second for "g".

2. A ball is launched at an angle of 56.5 degrees up from the horizontal, with a muzzle velocity of 7 meters per second, from a launch point which is 1 meters above the floor. How high will the ball be above the floor (in meters), when it is a horizontal distance of 2.6 meters away? Use 9.82 meters per second for "g".

3. A ball is launched at an angle of 35.8 degrees up from the horizontal, with a muzzle velocity of 7.1 meters per second, from a launch point which is 0.7 meters above the floor. How far horizontally (in meters) from the launcher will it land on the floor? Use 9.82 meters per second for "g".

4. For question #3, if you were asked to find the horizontal distance from the launcher when it is some distance ABOVE the floor (instead of when it HITS the floor), what about your previous equations/calculation would you have to change?

 a. Everything changes; you'll have to re-strategize this whole problem from the beginning. b. The only thing that changes is the final (or initial) y position. c. Nothing changes; the answer will be the same number as before. d. The only thing that changes is the launch angle.

Please explain the process in finding these! I can't understand if the initial velocity of vertical motion is separate from horizontal motion and it's tripping me up. The answers I got were 4.85, 4.27, 5.13, b

1.

When ball is launched perfectly horizontally, then

V0x = V0

V0y = Initial vertical velocity = 0 m/sec

ax = acceleration in horizontal direction = 0 m/sec^2

ay = acceleration in vertical direction = 9.81 m/sec^2

Now when ball travelled 1.2 m in vertical direction, then using 2nd kinematic equation:

h = V0y*t + (1/2)*ay*t^2

1.2 = 0*t + (1/2)*9.81*t^2

t = sqrt (2*1.2/9.81) = 0.495 sec

Now distance travelled in horizontal direction is given by:

d = V0x*t

V0x = V0 = d/t

given that d = 2.6 m, then

V0 = 2.6/0.495

V0 = Initial horizontal speed = 5.25 m/sec

Please ask rest of the questions as a new question, since they are not related.

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