Question

A stone is projected vertically upward with a velocity of 20 meters per second. 2 seconds later, a second stone is similarly projected with the same velocity. When the two stones meet, the second one is rising at a velocity of 10 meters per second. Neglecting air resistance, calculate the length of time the second stone is in motion before they meet. Also, find the velocity of the first stone when they meet. Take g equals to 10 meters per second.

Answer 1

To find the meeting point of two stones thrown vertically, apply projectile motion equations. The second stone is in motion for 1 second before they meet, and the velocity of the first stone is 0 m/s at the meeting point.

Step-by-step explanation:

The student asked about the meeting point of two stones thrown vertically upward with the same initial velocity but at different times. We need to calculate the length of time the second stone is in motion before they meet, as well as the velocity of the first stone upon meeting. We'll use the principles of projectile motion and assume a uniform acceleration due to gravity (g = 10 m/s2).

First, let's denote the time the first stone has been in motion when they meet as t and the time the second stone has been in the air as t - 2 seconds (since it started 2 seconds after the first stone).

The velocity of the second stone at the meeting point is given as 10 m/s. Applying the formula v = u + at (where u is the initial velocity, v is the final velocity, and a is the acceleration), we have:

10 m/s = 20 m/s + (-10 m/s2)(t - 2)

Rearranging, we find t = 3 seconds. So the second stone has been in motion for 1 second when they meet.

To find the velocity of the first stone, we use the same formula with t = 3 seconds:

v = 20 m/s + (-10 m/s2)(3)

Therefore, the velocity of the first stone is 0 m/s at the meeting point, indicating that it has reached the peak of its trajectory and is about to start descending.