Throwing a Pebble:Measuring Velocity and Acceleration

[chrhssupermanwl]

A Boy is standing at the edge of a cliff. He holds his hand over the edge of the cliff and throws a pebble directly up into the air with an initial speed of 11.6 m/s.
a) How long does it take for the pebble to reach maximum height?
b) What is the pebbles velocity when it hits the base of the cliff exactly 7.16 seconds after it was thrown? What is the pebble's speed at the same point in time?




average speed= distance/time; average velocity= (VF+VI)/2; average velocity= displacement in position/change in time; average acceleration= change in velocity/change in time; change in position= 1/2at^2

 

[Astronuc]

OK, the boy throws the ball vertically with an initial velocity of 11.6 m/s and the ball accelerates at a rate of -9.8 m/s².

What is the equation for v(t)?

Here is a handy reference - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

Specifically - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra1

 

[ogb p]

+VIt+Xo

I would approach this situation by first analyzing the given information and identifying the relevant equations and variables. From the given information, we know that the boy throws a pebble with an initial speed of 11.6 m/s and we are asked to determine the time it takes for the pebble to reach maximum height and its velocity and speed when it hits the base of the cliff after 7.16 seconds.

a) To determine the time it takes for the pebble to reach maximum height, we can use the equation for displacement in position: change in position = 1/2at^2 + VIt + Xo. In this case, the displacement at maximum height is zero, so the equation becomes 0 = 1/2at^2 + VIt + Xo. We know that the initial position (Xo) is at the edge of the cliff and the initial velocity (VI) is 11.6 m/s. Solving for time (t), we get t = 1.19 seconds. Therefore, it takes 1.19 seconds for the pebble to reach maximum height.

b) To determine the pebble's velocity and speed when it hits the base of the cliff after 7.16 seconds, we can use the equations for average velocity and average acceleration. The average velocity equation is VF = VI + at, where VF is the final velocity and a is the acceleration. The average acceleration equation is a = (VF-VI)/t. We know that at the base of the cliff, the final position is the same as the initial position (Xo), so the displacement is zero. Therefore, the average velocity at this point is also zero. Plugging in the given values, we get 0 = 11.6 + a(7.16). Solving for a, we get a = -1.62 m/s^2. This means that the pebble is experiencing a downward acceleration due to gravity. To determine the velocity, we can use the equation VF = VI + at, where VF is the final velocity, VI is the initial velocity, a is the acceleration, and t is the time. Plugging in the given values, we get VF = 11.6 + (-1.62)(7.16) = -0.4 m/s. Therefore, the pebble's velocity when it hits the base of the cliff is