Question
A cannon ball is shot out of a cannon buried 2m in the ground. The height of the cannonball can be approximated by the equation h=-5t^2+35t-2 where is the height of the cannonball in meters and t is the time in seconds.
a)how long will it take the cannonball to reach ground level, to the nearest tenth of a second
b)how long is the cannonball in the air to the nearest tenth of a second
c)find the maximum height of the cannonball and the time it takes to reach this height.
what steps would i need to do this
a)how long will it take the cannonball to reach ground level, to the nearest tenth of a second
b)how long is the cannonball in the air to the nearest tenth of a second
c)find the maximum height of the cannonball and the time it takes to reach this height.
what steps would i need to do this
Answers
To reach the ground, the height would be zero.
0 = -5t^2 + 35t - 2
You need to use the quadratic equation to solve for t, because this is not factorable.
b) Time up = time down
c) -b/2a will give you the max. height. -35/-10 = 3.5 meters
-
0 = -5t^2 + 35t - 2
You need to use the quadratic equation to solve for t, because this is not factorable.
b) Time up = time down
c) -b/2a will give you the max. height. -35/-10 = 3.5 meters
-
.o6
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