Author 
Topic 

dauerbach
Average Member
USA
13 Posts 
Posted  08/27/2007 : 20:30:12

If I could find my old book I could do this, but without it it has me stumped.
Solve algebraically:
A porjectile is launched upaward from ground level with an initial speed of 196 meter/sec. Use the position equation h = 4.9t^2 + vt where h = height, t = time,in seconds and v = initial velocity.
When will it return to the ground?
How high will it go? 


the_hill1962
Advanced Member
USA
1468 Posts 
Posted  08/28/2007 : 07:33:14

Putting in the 196m/s into your equation gives h = 4.9t + 196t To find the zeroes, you can factor to get h = 4.9t(t40) You see that h=0 when t=0 and t=40. To find the maximum height, you can use the power of calculus (probably beyond your need) or graph it on a graphing calculator (or by hand I suppose) and seeing where the max occurs. Or, you can use the fact that the max for an equation such as y=ax + bx + c occurs at the value x = b/2a. For your equation, a=4.9 b=196, so b/2a=20 This means t=20 at max. How high will that be? h = 4.9(20) + 196(20) = 1960



Subhotosh Khan
Advanced Member
USA
9116 Posts 
Posted  08/28/2007 : 11:00:07

quote: Originally posted by dauerbach
If I could find my old book I could do this, but without it it has me stumped.
Solve algebraically:
A porjectile is launched upaward from ground level with an initial speed of 196 meter/sec. Use the position equation h = 4.9t^2 + vt where h = height, t = time,in seconds and v = initial velocity.
When will it return to the ground?
How high will it go?
The vertex of a parabola (expressed as quadratic equation in 'x') will be at the midpoint between two roots.
This equation has roots at 0 and 40.
So the maximum height is reached at t = 20. 


HallsofIvy
Advanced Member
USA
78 Posts 
Posted  09/07/2007 : 10:40:11

h = 4.9t^2 + 196t Since when t= 0, h= 0 it is clear that "on the ground" corresponds to h= 0. The projectile will "hit the ground" when h= 0 again. Solve the equation 4.9t^2+ 196t= 0. One solution is t= 0. What is the other?
You can always find the maximum point of a parabola by COMPLETING THE SQUARE. But a sneaky short cut is to use the symmetry of the parabola. The projectile will be be at its highest point at exactly half the time it takes to hit the ground. Take half of the number you got from the first part, put it into the formula for h. 



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