The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21.5 m/s is
h = 3 + 21.5t − 4.9t2
after t seconds. (Round your answers to two decimal places.)
(a) Find the velocity after 2 s and after 4 s.
v(2) = m/s
v(4) = m/s
(b) When does the projectile reach its maximum height?
s
(c) What is the maximum height?
m
(d) When does it hit the ground?
s
(e) With what velocity does it hit the ground?
m/s
(a) To find the velocity after 2 seconds and after 4 seconds, we can take the derivative of the height equation with respect to time:
v(t) = d(h(t))/dt = d/dt (3 + 21.5t - 4.9t^2)
= 21.5 - 9.8t
Now we can substitute the values of t into the velocity equation:
v(2) = 21.5 - 9.8(2)
= 21.5 - 19.6
= 1.9 m/s
v(4) = 21.5 - 9.8(4)
= 21.5 - 39.2
= -17.7 m/s
Therefore, the velocity after 2 seconds is 1.9 m/s and the velocity after 4 seconds is -17.7 m/s.
(b) The projectile reaches its maximum height when the vertical velocity is zero. So, we set v(t) = 0 and solve for t:
21.5 - 9.8t = 0
Simplifying the equation:
9.8t = 21.5
t = 21.5 / 9.8
t ≈ 2.19 seconds
Therefore, the projectile reaches its maximum height at approximately 2.19 seconds.
(c) To find the maximum height, we substitute the value of t into the height equation:
h(max) = 3 + 21.5(2.19) - 4.9(2.19)^2
Simplifying the equation:
h(max) ≈ 3 + 47.085 - 21.288
h(max) ≈ 28.797 meters
Therefore, the maximum height reached by the projectile is approximately 28.797 meters.
(d) The projectile hits the ground when the height, h(t), is equal to zero. So, we set h(t) = 0 and solve for t:
3 + 21.5t - 4.9t^2 = 0
This is a quadratic equation, and we can solve it by either factoring or using the quadratic formula. Factoring is not feasible here, so we will use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Using the values from the height equation:
a = -4.9
b = 21.5
c = 3
t = (-(21.5) ± √((21.5)^2 - 4(-4.9)(3))) / (2(-4.9))
Simplifying the equation:
t ≈ (-21.5 ± √(462.25 + 59.04)) / (-9.8)
t ≈ (-21.5 ± √521.29) / (-9.8)
Using a calculator to find the square root:
t ≈ (-21.5 ± 22.83) / (-9.8)
There are two solutions for t:
t ≈ (-21.5 + 22.83) / (-9.8) ≈ 0.13 seconds (ignoring negative solution)
Therefore, the projectile hits the ground at approximately 0.13 seconds.
(e) To find the velocity at which it hits the ground, substitute the value of t into the velocity equation:
v(0.13) = 21.5 - 9.8(0.13)
= 21.5 - 1.274
≈ 20.23 m/s
Therefore, the projectile hits the ground with a velocity of approximately 20.23 m/s.
To answer the given questions, we will use the equation for the height of the projectile:
h = 3 + 21.5t - 4.9t^2
(a) To find the velocity after 2 seconds, we need to find the derivative of the height equation with respect to time:
v(t) = d/dt (3 + 21.5t - 4.9t^2)
Taking the derivative, we get:
v(t) = 21.5 - 9.8t
Now, substitute t = 2 into the equation:
v(2) = 21.5 - 9.8(2)
Simplifying, we have:
v(2) = 21.5 - 19.6
v(2) = 1.9 m/s
Therefore, the velocity after 2 seconds is 1.9 m/s.
To find the velocity after 4 seconds, substitute t = 4 into the equation:
v(4) = 21.5 - 9.8(4)
Simplifying, we have:
v(4) = 21.5 - 39.2
v(4) = -17.7 m/s
Therefore, the velocity after 4 seconds is -17.7 m/s.
(b) To find when the projectile reaches its maximum height, we need to find the time at which the velocity becomes 0.
Setting the velocity equation equal to 0:
0 = 21.5 - 9.8t
Solving for t, we have:
9.8t = 21.5
t = 21.5 / 9.8
t ≈ 2.19 s
Therefore, the projectile reaches its maximum height at approximately 2.19 seconds.
(c) To find the maximum height, we substitute t = 2.19 into the height equation:
h = 3 + 21.5(2.19) - 4.9(2.19)^2
Simplifying, we have:
h ≈ 3 + 47.085 - 10.88038
h ≈ 39.20 m
Therefore, the maximum height is approximately 39.20 meters.
(d) To find when the projectile hits the ground, we need to find the time at which the height becomes 0.
Setting the height equation equal to 0:
0 = 3 + 21.5t - 4.9t^2
Rearranging the equation, we have:
4.9t^2 - 21.5t - 3 = 0
This is a quadratic equation, and we can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our quadratic equation, a = 4.9, b = -21.5, and c = -3.
Substituting these values into the quadratic formula, we get:
t = (-(-21.5) ± √((-21.5)^2 - 4(4.9)(-3))) / (2(4.9))
Simplifying, we have:
t ≈ (-(-21.5) ± √(462.25 + 58.8)) / 9.8
t ≈ (21.5 ± √521.05) / 9.8
Using a calculator, we find two solutions: t ≈ 0.24 s and t ≈ 4.36 s.
Since the projectile is shot vertically upward, we are only interested in the positive solution.
Therefore, the projectile hits the ground at approximately 4.36 seconds.
(e) To find the velocity at which it hits the ground, substitute t = 4.36 into the velocity equation:
v(4.36) = 21.5 - 9.8(4.36)
Simplifying, we have:
v(4.36) = 21.5 - 42.728
v(4.36) = -21.23 m/s
Therefore, the velocity at which the projectile hits the ground is approximately -21.23 m/s.
The height (in meters) of a projectile shot vertically upward from a point 4 m above ground level with an initial velocity of 21.5 m/s is
h = 4 + 21.5t − 4.9t2
after t seconds. (Round your answers to two decimal places.)
(a) Find the velocity after 2 s and after 4 s.
v(2) =
1.9
m/s
v(4) =
-17.7
m/s
(b) When does the projectile reach its maximum height?
2.19
s
(c) What is the maximum height?
27.58
m
(d) When does it hit the ground?
109.64
s
(e) With what velocity does it hit the ground?
m/s