SOLUTION: A projectile is launched from ground level with an initial velocity of v0 feet per second. Neglecting air​ resistance, its height in feet t seconds after launch is given by s
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Question 1066609: A projectile is launched from ground level with an initial velocity of v0 feet per second. Neglecting air resistance, its height in feet t seconds after launch is given by s=−16t^2+v0t. Find the time(s) that the projectile will (a) reach a height of 96 ft and (b) return to the ground when v0 = 112 feet per second.
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
s=-16t^2+112t
96=-16t^2+112t
0=-16(t^2-7t+6), moving the 96 over and dividing everything by -16
(t-6)(t-1)=0
t=1, 6 seconds on the parabola.
It hits the ground when s=0 and -16t^2+112t=0; -16t(t-7)=0; t=7 seconds. t=0, of course, because it started there.
Answer by ikleyn(52777) (Show Source): You can put this solution on YOUR website!
.
On objects launched upward see the lessons
- Problem on a projectile moving vertically up and down
- Problem on an arrow shot vertically upward
- Problem on a ball thrown vertically up from the top of a tower
- Problem on a toy rocket launched vertically up from a tall platform
in this site.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic
"Projectiles launched/thrown and moving vertically up and down".
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