a ball is kicked in the air from the top of a cliff. the path the ball travels is given by the equation h(t)=-5t^2+17t+22 where h(t) is the height in metres and t is the time in seconds
a) what is the maximum height that the ball reaches
b)when will the ball hit the ground
Since the coefficient of t^2 is negative, the parabola opens downward and the max. point on the curve is the
vertex or max. ht.
V(h,k).
a. h = Xv = -b / 2a = -17 / -10 = 1.7.
Substitute 1.7 for t in the given Eq
and get:
h = k = Yv = 36.45m. = Height above cliff.
V(1.7,36.45).
b. H = 22 + 36.45 = 58.45m. = Ht. above
ground.
t(up) = (Vf - Vo) / g,
t(up) = (0 - 17) / -9.8 = 1.73s.
h = Vo*t + 0.5g*t^2 = 58.45m,
17t + 4.9*t^2 = 58.45,
4.9t^2 + 17t - 58.45 = 0,
Use Quadratic Formula to find t:
t(dn) = 2.13s.
T = t(up) + t(dn) = 1.73 + 2.13 = 3.86s
= Time in flight=Time to reach ground.
Correction:
h = Vo*t + 0.5*t^2 = 58.45m,
0 + 4.9t^2 = 58.45,
t^2 = 11.93,
t(dn) = 3.45s.
T = t(up) + t(dn) = 1.73 + 3.45 = 5.18s. = Time in flight = Time to reach ground.
To find the answer to the given question, we need to use the equation for the height of the ball at any given time, which is h(t) = -5t^2 + 17t + 22.
a) To find the maximum height that the ball reaches, we need to determine the vertex of the parabolic function. The vertex can be found using the formula t = -b / (2a), where t represents time, and a and b are the coefficients from the quadratic equation.
In our case, the equation can be rewritten as h(t) = -5t^2 + 17t + 22, so a = -5 and b = 17. Substituting these values into the formula, we get:
t = -17 / (2 * -5)
t = -17 / -10
t = 1.7 seconds
Now we can find the maximum height by substituting t = 1.7 back into the equation:
h(1.7) = -5(1.7)^2 + 17(1.7) + 22
h(1.7) = -14.45 + 28.9 + 22
h(1.7) = 36.45
Therefore, the maximum height that the ball reaches is 36.45 meters.
b) To find when the ball hits the ground, we set h(t) = 0 and solve for t:
-5t^2 + 17t + 22 = 0
To solve this quadratic equation, you can use factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
t = (-17 ± sqrt(17^2 - 4 * -5 * 22)) / (2 * -5)
After simplifying the equation, we get two possible solutions for t:
t = (-17 ± sqrt(289 + 440)) / -10
t = (-17 ± sqrt(729)) / -10
t = (-17 ± 27) / -10
So we have two possible values for t:
t1 = (27 - 17) / -10 = 1 second
t2 = (-27 - 17) / -10 = 4.4 seconds
Since time cannot be negative (t cannot be 4.4 seconds), we discard t2 as a non-relevant solution. Therefore, the ball will hit the ground after 1 second.
In conclusion:
a) The maximum height that the ball reaches is 36.45 meters.
b) The ball will hit the ground after 1 second.