1. Max. Speed = 100 mi/h.
100 mi / 3600 s * 5280 ft/mi = 146.7
ft/s.
d = Vo*t + 16t^2 = 20 ft,
146.7t + 16t^2 = 20,
16t^2 + 146.7t - 20 = 0,
Use Quad. Formula and get:
t = 0.1356s, and -9.3s.
Use the positive value:
t = 0.1356 s.
2. Vf^2 = Vo^2 + 2ad,
Vf^2 = (146.7)^2 + 2 * 32 * 20,
Vf^2 = 21521 + 1280,
Vf^2 = 22801,
Vf = 151 ft/s. = max. velocity.
3. Vmin^2 = Vo^2 + 2 * 32 * 0,
Vmin^2 = (146.7)^2 + 0,
Vmin = 146.7 ft/s = Slowest speed(h = 20 ft above gnd.
Pick a car and find its top speed. Assuming that your chosen car reaches its top speed at the top of a ramp, leaving the ramp vertically 20 feet about the ground: 1.) When will the car land back on the ground? 2.) When would the car be traveling the fastest? 3.) When would the car be traveling the slowest? 4) based on the speed of the car and how long it was in the air, estimate how high into the air the car would have reached. h(t)=-16tsquared-Vo initial velocity ho initial height. height in feet velosity in ft./sec.
2 answers
thank you henry you rock