Heather, here you go:
#15. Use the Second Kinematics Equation. Initial velocity is zero. so the equation becomes Delta y = 1/2at^2. Solve this for t and you find that t is proportional to the square root of delta y. So the package that is dropped from a height of 4H takes 2 times as much time to hit the ground as the package at height h. Answer B.
#17. Use the Third Kinematics Equation (there is no time in the problem). The final velocity at the max height is zero, so the equation, solved for delta y is: delta y = (v initial)^2/2a. The max height depends on the square of the initial velocity - so the arrow fired at 4 times the initial velocity of the other one rises to a height 16 times that one. Answer D.
#20. Use the second kinematics equation with the initial velocity and position equal to zero. Solve it for a.
a = 2x/t^2. Plug in the position and time at any of the points and you get a = 4 m/s^2. If you take the last point where x = 50 and t = 5 you get: a = 2(50)/5^2 = 4. Answer C.

## 4 Comments

Christopher Crocker •4 years, 9 months ago • login to replyRosanna Satterfield •4 years, 6 months ago • login to replyHeather Vinick •2 years, 6 months ago • login to replyJohn Ennis •2 years, 6 months ago • login to reply