SOLUTION: A ball is thrown vertically upward from the top of a building 102 feet tall with an initial velocity of 60 feet per second. The distance s (in feet) of the ball from the ground af
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Question 299015: A ball is thrown vertically upward from the top of a building 102 feet tall with an initial velocity of 60 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s(t)=-16t^2+60t+102
What is the distance of the ball from the ground at time t=0?
After how many seconds does the ball strike the ground?
After how many seconds will the ball pass the top of the building on its way down?
You can put this solution on YOUR website! A ball is thrown vertically upward from the top of a building 102 feet tall with an initial velocity of 60 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s(t)=-16t^2+60t+102
What is the distance of the ball from the ground at time t=0?
Set t=0 and plug into formula:
s(t)=-16t^2+60t+102
s(0)=-16*0^2+60(0)+102
s(0)= 102 feet
.
After how many seconds does the ball strike the ground?
Solve for t when s(t)=-102 (that's because we are 102 feet above the ground)
s(t)=-16t^2+60t+102
-102=-16t^2+60t+102
0=-16t^2+60t+204
0=-16t^2+60t+204
0=-4t^2+15t+51
.
Solve using the quadratic formula. Doing so yields:
x = {-2.16, 5.91}
Throw away the negative solution leaving:
t = 5.91 seconds
.
After how many seconds will the ball pass the top of the building on its way down?
Solve for t when s(t)=0
s(t)=-16t^2+60t+102
0=-16t^2+60t+102
0=-8t^2+30t+51
Solve using the quadratic formula (Details below). Doing so yields:
x = {-1.27, 5.02}
Throw away the negative solution leaving:
t = 5.02 seconds
.
Details of quadratic for part c:
