Problema Solution
A spittlebug jumps into the air with an initial velocity of 10 feet per second.
A. Write an equation that gives the height of the spittlebug as a function of the time (in seconds) since it left the ground.
B. The spittlebug reaches it's maximum height after 0.3125 second. How high can it jump?
Answer provided by our tutors
A.
y = (1/2)a*t^2 + v0t + y0
a = - 32 ft/s^2 acceleration due to gravity
v0 = 10 ft/s the initial velocity
y0 = 0 ft initial height
t = time in seconds
y = the height as function of the time t
y = - (1/2)*32*t^2 + 10t + 0
y = - 16*t^2 + 10t
B.
y max is reached after 0.3125 that is for t = 0.3125 s
y max = - 16*(0.3125)^2 + 10*0.3125
y max = 1.5625 ft
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