Question

To understand how the trajectory of an object depends on its initial velocity, and to understand how air resistance affects the trajectory.

For this problem, use the PhET simulation Projectile Motion. This simulation allows you to fire an object from a cannon, see its trajectory, and measure its height, range, and hang time (the amount of time in the air).

Start the simulation by selecting the option labeled "Intro". Press the red Fire icon to launch an object. You can choose the object by clicking on one of the objects in the scroll-down menu at top right. To adjust the cannon barrel's angle, click and drag on it. You can also adjust the Initial Speed of the object using the toolbar at the bottom left. Clicking Air Resistance displays the drag coefficient for the given object and enables air resistance during motion. To measure the height, range, or hang time, drag the box that contains these quantities from the top of the screen to the various points of interest along the object's trajectory (use the crosshairs to locate the exact point of interest).

To understand how the trajectory of an object depends on its initial velocity, and to understand how air resistance affects the trajectory.

For this problem, use the PhET simulation Projectile Motion. This simulation allows you to fire an object from a cannon, see its trajectory, and measure its height, range, and hang time (the amount of time in the air).

When the pumpkin is shot straight upward with an initial speed of 14 m/s, what is the maximum height above its initial location?

Notice that this value could be determined from the kinematics equation. Since you found it takes 2.9 s for the pumpkin to reach the ground, it must take 1.43 s to reach the maximum height, which gives y(t=1.43s)=(14m/s)(1.43s)−(1/2)(9.8m/s2)(1.43s)2 = 10m.

Part C

If the initial speed of the pumpkin is doubled, how does the maximum height change? (Note: for this part, as well as later parts, you will need to use the zoom in and out buttons to see the full trajectories)

Answer 1

if the initial speed (u) of pumpkin is 14 m/s

final speed i.e. speed at the highest point (v) = 0

let, t be the time taken by pumpkin to reach the maximum height

therefore, v = u + gt

or, 0 = 14 + (-9.8) t

or, -9.8 t = -14

or, t = 14/ 9.8 = 1.43 s

now maximum height (y₁) = ut + 1/2 gt² = (14 x 1.43) + 1/2 x (-9.8) x (1.43)²

or, y₁ = 20.02 - 10.02 = 10 m

if the initial speed (u) of pumpkin is doubled i.e. 28 m/s

final speed i.e. speed at the highest point (v) = 0

let, t be the time taken by pumpkin to reach the maximum height

therefore, v = u + gt

or, 0 = 28 + (-9.8) t

or, -9.8 t = -28

or, t = 28 / 9.8 = 2.86 s

now maximum height (y₂) = ut + 1/2 gt² = (28 x 2.86) + 1/2 x (-9.8) x (2.86)²

or, y₂ = 80.08 - 40.08 = 40 m

or, y₂ = 4 x 10 m = 4 y₁

hence, the maximum height becomes 4 time if the initial speed of the pumpkin is doubled .