When we estimate distances from velocity data, it is sometimes necessary to use times to, t1, t2, t3, ... that are not equally spaced. We can still estimate distances using the time periods At; = t; - t; - 1. For example, a space shuttle was launched on a mission, the purpose of which was to install a new motor in a satellite. The table provided gives the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. Use these data to estimate the height, h, above Earth's surface of the space shuttle, 62 seconds after liftoff. (Give the upper approximation available from the data.) h = ft Event Time (s) Velocity (ft/s) Launch Begin roll maneuver 10 180 End roll maneuver 15 319 Throttle to 89% 20 442 Throttle to 67% 32 742

A ball is thrown upward from the top of a building with a velocity of 14.7 meters / sec. and after reaching its maximum height, it begins to descend in free fall until it reaches the ground (down on the road). If we take the top of the building as a reference: a) Determine the time (tymax.) at which the ball reaches its maximum height and the value of said maximum height (ymax-). b) Determine the position (y) and the velocity (v,) of the ball for a time equal to 2 seconds. For the 2-second time, is the ball above the top of the building or below the top of the building? Is the ball going up or down? c) Determine the position (y) and the velocity (v,) of the ball for a time equal to 4 seconds. For the 4-second time, is the ball above the top of the building or below the top of the building? Is the ball going up or down? d) Determine the velocity (v,) of the ball at a height of 6 meters above the top of the building. Explain your answer in terms of the signs obtained in your result.

Bob hits a tennis ball upward with a velocity of Vo= 30 y m/s from a height of 1 meter above the ground. a) Find the velocity of the ball after 2.5 seconds (Answer: 5.5y m/s  ) b) Find the position of the ball at this time (Answer: 45.4y m ) c) Plot acceleration, velocity, and position vs time.

dog sees a squirrel 13.6 m away. The dog runs toward the squirrel at 5.3 m/s. The squirrel was not paying attention and didn't noticethe dog until the dog was only 2.1 m away. If the tree is 11.67 m away from the squirrel (in the opposite direction of the dog), what is theminimum speed the squirrel will need to run at to get away (ie get to the tree at least at the same time the dog does)?