a. V^2 = Vo^2 + 2g*hc.
hc = (Vf^2-Vo^2)/2g = Ht. above the cliff.
b. h max = ho + (Vf^2-Vo^2)/2g
h max = 58.8 + (0-49^2)/(-19.6) = 181.3 m. Above gnd.
c. V = Vo + g*Tr = 0.
49 -9.8Tr = 0, Tr = 5 s. = Rise time.
h max = 0.5g*Tf^2 = 181.3.
4.9Tf^2 = 181.3, Tf = 6.08 s. = Fall time.
Tr+Tf = 5 + 6.08 = 11.08 s. = Time to hit gnd.
d. V^2 = Vo^2 + 2g*h.
V^2 = 0 + 19.6(181.3-100) = 1593.48, V = 39.9 m/s.
V = Vo + g*t = 39.9 m/s.
0 + 9.8t = 39.9, t = 4.07 s.
e. h = Vo*t + 0.5g*t^2.
V0 = 49 m/s, t = 2 s., g = -9.8 m/s^2, h = ?.
an arrow is shot straight up from a cliff 58.8 meters above the ground with an initial velocity of 49 meters per second
a. write an equation that represents this situation
b. Transform the equation to the format that can most easily be used to find the maximum height. What is the height?
c. Transform the equation to a different format to determine how many seconds it would take for the arrow to hit the ground.
d. Show how you would use the equation in part (a) to determine approximately when the arrow would be 100m above the base of the cliff
e. Show how you could determine the height of the arrow after 2 secs
f. Sketch the graph of the equation
2 answers
510