Boris throws a ball vertically upward from the top of a cliff. The height of the hall above the base of the cliff is approximated by the model h=65+10t-5t^2, where h is the height in metres and t is the time in seconds.
a. How high is the cliff?
b. How long does it take for the ball to reach a height of 5om above the base of the cliff?
C. After how many seconds does the ball hit the ground?
For (a) I got 65metres, For (b) I got 3secs BUT I DO NOT KNOW THE ANSWER TO C .
PLS HELP
a is obvious
b 5 = 65 + 10 t -5 t^2
5 t^2 - 10 t - 60 = 0
t^2 -2 t -12 = 0
t = [ 2 +/- sqrt (4+48)]/2
= 1 +/- sqrt 13
= 4.61 seconds
c
0 = 65 + 10 t - 5 t^2
t^2 -2 t -13 = 0
t = [ 2 +/- sqrt(4+52)]/2
= 1 +/- sqrt(14)
=4.74 sec
To find the answer for part (c), we need to determine when the ball hits the ground. In this case, the ball hits the ground when the height (h) becomes zero.
The equation for the height of the ball is h = 65 + 10t - 5t^2.
To find when the ball hits the ground, we need to find the value of t that makes h = 0. So we can set up the equation:
0 = 65 + 10t - 5t^2
To solve this quadratic equation, we can use factoring, the quadratic formula, or graphical methods. In this case, let's use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
Here, a = -5, b = 10, and c = 65.
t = (-10 ± sqrt(10^2 - 4(-5)(65))) / (2*(-5))
t = (-10 ± sqrt(100 + 1300)) / (-10)
t = (-10 ± sqrt(1400)) / (-10)
Now, we can simplify further:
t = (-10 ± sqrt(4 * 25 * 7)) / (-10)
t = (-10 ± 2 * 5 * sqrt(7)) / (-10)
t = (-10 ± 10sqrt(7)) / (-10)
t = 1 ± sqrt(7)
Since we are looking for the time it takes for the ball to hit the ground, we only consider the positive value:
t = 1 + sqrt(7)
Therefore, the ball hits the ground approximately 1 + sqrt(7) seconds after it was thrown.