The height h, in metres, of a ball thrown from a cliff can be modelled by the function with the rule h(t) = 20+15t-5t^2, where t less than/equal to 0 is the time measured in seconds.
a) find the initial height of the ball
b) find the time at which the ball hits the ground
Thanks would be highly appreciated!
initial ----> t = 0
So what do you get when you replace t with 0 in
h(t) = 20 + 15t - 5t^2 ?
When it hits the ground, isn't the height zero?
0 = 20 + 15t - 5t^2
divide by -5
t^2 - 3t - 4 = 0
solve the quadratic using whatever method you know, make sure to reject
the negative value of t
( it factors, you should be able to see the factors in your head)
To find the initial height of the ball, we need to determine the value of h(0).
a) The function h(t) gives the height of the ball at any given time t. To find the initial height of the ball, we substitute t = 0 into the function:
h(0) = 20 + 15(0) - 5(0)^2
h(0) = 20
Therefore, the initial height of the ball is 20 meters.
b) To find the time at which the ball hits the ground, we need to find the value of t when h(t) is equal to 0 (since the height of the ball will be 0 when it hits the ground).
Let's set h(t) = 0 and solve for t:
0 = 20 + 15t - 5t^2
Rearranging the equation:
5t^2 - 15t - 20 = 0
This is a quadratic equation, which can be solved using factoring, completing the square, or the quadratic formula. In this case, let's solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 5, b = -15, and c = -20. Plugging these values into the quadratic formula:
t = (-(-15) ± √((-15)^2 - 4(5)(-20))) / (2(5))
t = (15 ± √(225 + 400)) / 10
t = (15 ± √625) / 10
t = (15 ± 25) / 10
We have two possible solutions:
1) t = (15 + 25) / 10 = 40 / 10 = 4
2) t = (15 - 25) / 10 = -10 / 10 = -1
Since t represents time, we discard the negative solution. Therefore, the ball hits the ground after 4 seconds.
To summarize:
a) The initial height of the ball is 20 meters.
b) The ball hits the ground after 4 seconds.