A projectile is launched straight up at 54.5 m/s from a height of 88 m, at the edge of a sheer cliff. The projectile falls, just missing the cliff and hitting the ground below.
(a) Find the maximum height of the projectile above the point of firing.
151.5
I calculated and found the maximum height just struggling for the other parts
(b) Find the time it takes to hit the ground at the base of the cliff.
s
(c) Find its velocity at impact.
m/s
Hi = 88
Vi = +54.5
a = -9.81
v = Vi - 9.81 t
v = 0 at top
so at top t = 54.5 / 9.81 = 5.56 seconds
then h = Hi + Vi t - 4.9 t^2
= 88 + 54.5 (5.56) - 4.9 (5.56)^2 = 88 + 303 -151 = 240 meters which is 240-88 = 152 meters above firing point
when does h = 0 ?
0 = 88 + 54.5 t - 4.9 t^2
solve quadratic for t
https://www.mathsisfun.com/quadratic-equation-solver.html
roots are t = -1.43 or t = +12.5
the negative root is when it would have been on the ground before the firing
use t = 12.5 seconds total in air
Then again v = Vi - 9.81 t
= 54.5 - 9.81 (12.5)
a. Vo^2 + 2g*h = V^2.
54.5^2 + (-19.6)h = 0,
h = 151.5 m.
b. ho + h = 88 + 151.5 = 239.5 m. above gnd.
0.5*g*Tf^2 = 239.5.
0.5*9.8*Tf^2 = 239.5,
Tf = 7 s. = Fall time.
Vo + g*Tr = V.
54.5 + (-9.8)Tr = 0,
Tr = 5.6 s. = Rise time.
T = Tr + Tf = 7+ 5.6 = 12.6 s. = Time to reach gnd.
c. V^2 = Vo^2 + 2g*h = 0 + 19.6 *239.5 = 4694.
V = 68.5 m/s.
To solve these problems, we can use the equations of motion for projectile motion. The equations we will use are:
1. Vertical displacement (y) = initial vertical velocity (v₀) * time (t) - (1/2) * acceleration due to gravity (g) * time squared (t²)
2. Final vertical velocity (v) = initial vertical velocity (v₀) - acceleration due to gravity (g) * time (t)
Let's solve the problems step by step:
(a) Finding the maximum height of the projectile above the point of firing:
To find the maximum height, we can use the equation for vertical displacement. At the maximum height, the final vertical velocity will be zero, so the equation becomes:
0 = (54.5 m/s) * t - (1/2) * (-9.8 m/s²) * t²
Simplifying the equation, we have:
-4.9t² + 54.5t = 0
Factoring out t, we get:
t * (-4.9t + 54.5) = 0
Solving for t, we have two solutions:
t₁ = 0 (Ignoring this as it represents the time at the beginning of the motion)
t₂ = (54.5 m/s) / (4.9 m/s²) ≈ 11.12 s
Substituting t = 11.12 s into the equation for vertical displacement, we find:
y = (54.5 m/s) * (11.12 s) - (1/2) * (-9.8 m/s²) * (11.12 s)²
≈ 151.5 meters
Therefore, the maximum height of the projectile above the point of firing is approximately 151.5 meters.
(b) Finding the time it takes to hit the ground at the base of the cliff:
To find the time it takes to hit the ground, we need to find the time when the vertical displacement becomes equal to the initial height (88 meters). Using the same equation for vertical displacement as before, we have:
88 = (54.5 m/s) * t - (1/2) * (-9.8 m/s²) * t²
Simplifying the equation, we have:
-4.9t² + 54.5t - 88 = 0
Solving this quadratic equation, we find:
t ≈ 9.64 s
Therefore, it takes approximately 9.64 seconds for the projectile to hit the ground at the base of the cliff.
(c) Finding its velocity at impact:
To find the velocity at impact, we can use the equation for final vertical velocity. At impact, the time will be equal to the time found in part (b), which is approximately 9.64 seconds. Substituting the values into the equation, we have:
v = (54.5 m/s) - (-9.8 m/s²) * (9.64 s)
= 54.5 m/s + 94.832 m/s
≈ 149.332 m/s
Therefore, the velocity at impact is approximately 149.332 m/s in magnitude.
To find the maximum height of the projectile above the point of firing, you can use the kinematic equation for projectile motion.
First, let's find the time it takes for the projectile to reach its peak height. When the projectile reaches its peak, its vertical velocity will be zero. You can use the following equation to find the time:
v_f = v_i + at
Here, v_f is the final velocity (which is 0 m/s at the peak), v_i is the initial velocity (which is 54.5 m/s in the upward direction), a is the acceleration (which is the acceleration due to gravity, approximately -9.8 m/s^2), and t is the time.
Rearranging the equation to solve for time, we have:
t = (v_f - v_i) / a
Substituting the given values:
t = (0 - 54.5) / -9.8
t ≈ 5.56 seconds (rounded to two decimal places)
Now, to find the maximum height, we can use the kinematic equation for vertical displacement:
d = v_i * t + (1/2) * a * t^2
In this equation, d represents the displacement (which is the maximum height), v_i is the initial velocity in the upward direction, t is the time calculated earlier, and a is the acceleration due to gravity.
Substituting the values:
d = 54.5 * 5.56 + (1/2) * (-9.8) * (5.56)^2
d ≈ 151.5 meters (rounded to one decimal place)
So, the maximum height of the projectile above the point of firing is approximately 151.5 meters.
Now, let's move on to finding the time it takes for the projectile to hit the ground at the base of the cliff.
To find the total time of flight, we need to consider both the upward and downward motion of the projectile. The time it takes for the projectile to reach its peak is already calculated as 5.56 seconds. The total time of flight will be twice this value since the upward and downward paths are symmetrical.
Therefore, the total time of flight is approximately 2 * 5.56 = 11.12 seconds.
For part (c), to find the velocity at impact, we can use the following equation:
v_f = v_i + at
Here, v_f is the final velocity (which is the velocity at impact), v_i is the initial velocity (which is 54.5 m/s in the downward direction), a is the acceleration (which is the acceleration due to gravity, approximately -9.8 m/s^2), and t is the time of flight calculated earlier.
Substituting the values:
v_f = 54.5 - 9.8 * 11.12
v_f ≈ -60.84 m/s (rounded to two decimal places)
Since the velocity is negative, it indicates that the projectile is moving downward.
Therefore, the velocity at impact is approximately -60.84 m/s (rounded to two decimal places).