VintageKitKat
Answered

Please help me, i'm trying to graduate!!!
A ball is thrown from a height of 151 feet with an initial downward velocity of 15/fts . The ball's height h (in feet) after t seconds is given by the following.


h=151−15t-16t^2


How long after the ball is thrown does it hit the ground? Round your answer(s) to the nearest hundredth.

Verified Answer :

jedster55

If you use this equation then you say that the ground is h=0 and solve as a quadratic.
The quadratic formula is (-b±√(b^2-4ac))/2a when an equation is in the form ax^2 + bx + c
So the equation you have been given would be -16t^2-15t-151 = 0
This equation has no real roots which leads me to believe it is incorrect.
This is probably where your difficulty is coming from, it's a mistake.

The equation is formed from S=ut+(1/2)at^2+(So) where (So) is the initial height and S is the height that you want to find.
In this case you want S = 0.
If the initial height is +151 and the initial velocity and acceleration are downwards (negative) and the initial velocity (u) is -15 and the initial acceleration is -32 then you get the equation S=-15t-16t^2+151

Solving this using the quadratic formula gives you t = 2.64 or t = -3.58
Obviously -3.58s can't be the answer so you're left with 2.64 seconds.

Hope this makes sense.

Answer:

2.64 seconds

Step-by-step explanation:

To find when the ball will hit the ground, you just need to use the equation of h with the value h = 0 (the ball is in the ground), and then find the value of t:

h = 151 − 15t - 16t^2

0 = 151 − 15t - 16t^2

To solve this quadratic equation, you can use Bhaskara formula:

Delta = b2 - 4ac = (-15)^2 - 4(-16)151 = 9889

sqrt(Delta) = 99.4435

t1 = [-b + sqrt(Delta)] / 2a = (15 + 99.4435) / (-32) = -3.5764 seconds (not useful result, because there is not a negative time)

t2 = [-b - sqrt(Delta)] / 2a = (15 - 99.4435) / (-32) = 2.6389 seconds

Rounding to the nearest hundredth: t = 2.64 seconds

Related Questions