Answer in Physics for arte08 #227566
A stone is thrown vertically upward from the top of the building. If the equation of the motion of the stone is ( ) = −5 2 + 30 + 200 , where is the directed distance from the ground in meters and is in seconds.
a. What is the height of the building?
b. What is the average velocity on the time interval [1,3]
c. What is the instantaneous velocity at time t=1?
d. at what time the stone will hit the ground?
e. What is the instantaneous velocity of the stone upon impact?
a. At “t = 0” the value of “s(t)” gives the height of the building. Thus:
“h = s(0) = 200m”
b. The distance from the ground at “t = 1” is:
“s(1) = -5cdot 1^2 + 30cdot 1 + 200 = 225m”
The distance from the ground at “t = 3” is:
“s(3) = -5cdot 3^2 + 30cdot 3 + 200 = 245m”
The average velocity is:
“v_{av} = dfrac{s(3) – s(1)}{3s – 1s} approx 6.7m/s”
c. The instantaneous velocity is:
“v(t) = s'(t) = -10t+30\nv(1) = -10+30 = 20m/s”
d. The stone will hit the ground when “s(t) = 0.” Solving the following quadratic equation for “t”, obtain:
“-5t^2 + 30t + 200 = 0\nt = 10s”
c. The instantaneous velocity at this time is:
“v(10) = -10cdot 10 + 30 = -70m/s”
