SOLUTION: A projectile is launched upward from the ground level with an initial speed of 98m/s. How high will it go? When will it return to the ground?

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Question 53174This question is from textbook Algebra and Trigonometry Structure and Method
: A projectile is launched upward from the ground level with an initial speed of 98m/s. How high will it go? When will it return to the ground? This question is from textbook Algebra and Trigonometry Structure and Method

Found 2 solutions by venugopalramana, Nate:
Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website!
A projectile is launched upward from the ground level with an initial speed of 98m/s. How high will it go? When will it return to the ground?
THE FORMULAE ARE ......
V=U-GT.....
V^2-U^2 = -2GS
WHERE U = INITIAL VELOCITY = 98 M/SEC
V = FINAL VELOCITY = 0 M/SEC
G = ACCELERATION DUE TO GRAVITY = 9.8 M/SEC^2
S = DISTANCE TRAVELLED IN M .
0 - 98^2 = -2*9.8S
S= 98*98/(9.8*2)= 490 M .....IT WILL GO UP TO 490 M . HT.
0 = 98 - 9.8 T
T=98/0.8 =10 SEC...
IT WILL TAKE SAME TIME TO COME DOWN
HENCE IT WILL HIT THE GROUND AFTER 10+10 = 20 SEC

Answer by Nate(3500) (Show Source):
You can put this solution on YOUR website!
y = -16x^2 + 98x
0 = -16(x^2 - 6.125x)
0 = (x - 3.0625)^2
9.37890625 = (x - 3.0625)^2
+- 3.0625 = x - 3.0625
3.0625 +- 3.0625 = x
6.125 seconds
Mean: (0 + 6.125)/2 = 3.0625
y = -16(3.0625)^2 + 98(3.0625)
y = -16(9.890625) + 98(3.0625)
y = -150.0625 + 300.125
y = 150.0625m