Let's break down the problem shown in the image:

Problem:

• Original graph: It displays the vertical position vs. time for an object launched vertically from the ground.
• Question: It asks which graph correctly represents the vertical position of the object if it were launched at the same angle but with twice the initial speed.

Key Concept:

When an object is launched with twice the initial speed:

• The initial vertical velocity is doubled.
• The maximum height achieved by the object will be four times higher, because height is proportional to the square of the velocity (using the formula $h = \frac{v_0^2}{2g}$).
• The total time of flight will also double, since time to reach the maximum height is proportional to velocity ($t = \frac{v_0}{g}$).

Analyzing the Answer (Graph A):

In the graph for option A:

• The maximum height is around 70 m, which is roughly four times the original height (the original height was about 17-18 m).
• The total time of flight appears to be 10 seconds, which is double the original 5 seconds.

Since both the height and time of flight are consistent with the expected changes for an object launched at twice the initial speed, Graph A is correct.

Your answer is correct.

Do you want further details or have more questions?

Here are 5 related questions you might want to explore:

1. What are the kinematic equations used to describe vertical motion under gravity?
2. How does doubling the angle of launch affect the range and maximum height of the projectile?
3. What would happen if the speed is tripled instead of doubled?
4. How does air resistance change the motion of a projectile in comparison to the ideal case?
5. How can we calculate the horizontal displacement if we launch a projectile at an angle?

Tip: Always analyze the effect of each variable separately (velocity, time, angle) in projectile motion to better understand the result.