A boy uses a slingshot to launch a pebble straight up into the air. The pebble reaches a height of 29.0 m above the launch point 3.3 seconds later. Assume air resistance is negligible.

a) What was the pebble's initial speed (just after leaving the slingshot)?

b) How much time did it take for the pebble to first reach a height of 14.5 m above its launch point?

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To find the initial velocity and acceleration of the pebble, we can use the kinematic equations.

We know that the final position (height) of the pebble is 29.0 m, the time it takes to reach that height is 3.3 seconds, and the initial velocity (when the pebble was launched) is unknown. Let's use the following equation to solve for the initial velocity:

Δy = v0*t + (1/2)*a*t^2

In this equation, Δy represents the change in position (29.0 m), v0 is the initial velocity we want to find, t is the time (3.3 seconds), and a is the acceleration (which we assume to be constant and equal to the acceleration due to gravity, -9.8 m/s^2).

Rearranging the equation, we have:

v0 = (Δy - (1/2)*a*t^2) / t

Substituting the known values:

v0 = (29.0 m - (1/2)*(-9.8 m/s^2)*(3.3 s)^2) / 3.3 s

Simplifying the equation:

v0 = (29.0 m + 51.048 m) / 3.3 s

v0 = 80.048 m / 3.3 s

v0 ≈ 24.26 m/s

Therefore, the initial velocity of the pebble when it was launched was approximately 24.26 m/s.

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