Word Problem: Find max height, landing time of projectile

[kvaughn_75681]

Initial velocity of a projectile is 230 feet per second then its height h is feet is a function of time t in seconds given by h(t) = -16t^2+230t. I am to find the maximum height of the projectile and the time when the projectile hits the ground.

The maximum height is -b/2(a) which is -230/2(-16) which is 198 feet. I think that is correct.

The time when the projectile hits the gound is written: -16t^2+230t=0. I think that I subtract 230t from both sides and then divide by -16, but I am not sure.

Any help in the right direction would be appreciated.

 

[jwpaine]

Complete the square!

\(\displaystyle \L -16t^2+ 230t = 0\)

\(\displaystyle \L t^2 -\frac{115}{8}t = 0\)

add \(\displaystyle \L (\frac{1}{2}b)^2\) to both sides

\(\displaystyle \L t^2 -\frac{115}{8}t + \frac{13225}{256} = \frac{13225}{256}\)

simplify LHS

\(\displaystyle \L (t - \frac{115}{16})^2 = \frac{13225}{256}\)

squareroot both sides

\(\displaystyle \L t = \frac{115}{16} (+/-) \frac{115}{16}\)

t = 0 or t = \(\displaystyle \L \frac{115}{8}\)

 

[TchrWill]

Re: Word Problem: Find max height, landing time of projectil

Initial velocity of a projectile is 230 feet per second then its height h is feet is a function of time t in seconds given by h(t) = -16t^2+230t. I am to find the maximum height of the projectile and the time when the projectile hits the ground.

The maximum height is -b/2(a) which is -230/2(-16) which is 198 feet. I think that is correct.

The time when the projectile hits the gound is written: -16t^2+230t=0. I think that I subtract 230t from both sides and then divide by -16, but I am not sure.

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Assuming the projectile is fired vertically:
Vo = initial velocity
Vf = final velocity
g = acceleration due to gravity = 32.2 ft./sec.^2t
t1 = time to maximum height
t2 = time from maximum height to ground = t1
h = maximum height.

The time to maximum height is derived from Vf = Vo - gt. Since Vf = 0, t = 230/32.2 = 7.142 sec.

The maximum height derives from h = Vot1 - gt1^2/2 = 230(7.142) - 32.2(7.142)^2/2 = 821 feet.

The time to ground is 2t1 = 14.284 sec.

 

[skeeter]

Re: Word Problem: Find max height, landing time of projectil

Initial velocity of a projectile is 230 feet per second then its height h is feet is a function of time t in seconds given by h(t) = -16t^2+230t. I am to find the maximum height of the projectile and the time when the projectile hits the ground.

The maximum height is -b/2(a) which is -230/2(-16) which is 198 feet. I think that is correct.

no ... t = -230/(-32) = 7.1875 sec

h(7.1875) = -16(7.1875)² + 230(7.1875) = approx 827 ft

The time when the projectile hits the gound is written: -16t^2+230t=0. I think that I subtract 230t from both sides and then divide by -16, but I am not sure.

you can do one of two things ... first (and easiest), double the time it took to get to its max height ... 2(7.1875) = 14.375 sec

second method ... set h(t) = 0 and solve for t

-16t² + 230t = 0

-2t(8t - 115) = 0

t = 0 or t = 115/8 = 14.375

Any help in the right direction would be appreciated.