A ball is thrown from a cliff and follows the path given by h= -5t2+10t+65 where h is the height in metres and t is the time in seconds.
Determine the height of the cliff.
Determine the length of time the ball spends above 66 m. Answer to two decimal places.
To determine the height of the cliff, we need to find the value of h when t=0.
h = -5(0)^2 + 10(0) + 65
h = 0 + 0 + 65
h = 65
Therefore, the height of the cliff is 65 meters.
To determine the length of time the ball spends above 66 meters, we need to set h = 66 and solve for t.
66 = -5t^2 + 10t + 65
0 = -5t^2 + 10t - 1
Using the quadratic formula, we can solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
t = (-10 ± √(10^2 - 4(-5)(-1))) / (2(-5))
t = (-10 ± √(100 - 20)) / (-10)
t = (-10 ± √80) / (-10)
t = (-10 ± 2√5) / (-10)
t = (10 ± 2√5) / 10
t = 1 ± √5
Since t cannot be negative in this context, we only consider the positive solution:
t = 1 + √5
t ≈ 3.24 to two decimal places.
Therefore, the ball spends approximately 3.24 seconds above 66 meters.
Determine the length of time the ball spends above 66 m. Answer to two decimal places
To determine the length of time the ball spends above 66 m, we need to find the values of t when h = 66.
66 = -5t^2 + 10t + 65
0 = -5t^2 + 10t + 1
Using the quadratic formula, we can solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
t = (-10 ± √(10^2 - 4(-5)(1))) / (2(-5))
t = (-10 ± √(100 + 20)) / (-10)
t = (-10 ± √120) / (-10)
t = (-10 ± 2√30) / (-10)
t = (10 ± 2√30) / 10
t = 1 ± √30
Since t cannot be negative in this context, we only consider the positive solution:
t = 1 + √30
t ≈ 6.45 to two decimal places.
Therefore, the ball spends approximately 6.45 seconds above 66 meters.
Determine the maximum height of the ball and the time that it reaches maximum height.
To determine the maximum height of the ball, we can find the vertex of the parabolic function h = -5t^2 + 10t + 65. The maximum height occurs at the vertex of the parabola.
The t-coordinate of the vertex can be found using the formula t = -b / (2a), where a = -5 and b = 10:
t = -10 / (2(-5))
t = -10 / (-10)
t = 1
To find the corresponding height, we substitute the value of t back into the equation:
h = -5(1)^2 + 10(1) + 65
h = -5 + 10 + 65
h = 70
Therefore, the maximum height of the ball is 70 meters.
The time that the ball reaches the maximum height is 1 second.