An object is launched at an angle of 78º above the horizontal with an initial speed of 120 m/s. Find (a) the maximum altitude reached by the object, (b) its total time of flight, and (c) its horizontal range
V1= 120m/s[78o].
X1 = 120*Cos78 = 24.9 m/s.
Y1 = 120*sin78 = 117.4 m/s.
a. Y2^2 = Y1^2 + 2G*h.
0 = 117.4^2 - 19.6h, h = ?.
b. V2 = V1 + g*Tr.
0 = 117.4 - 9.8Tr, Tr = 12 s. = Rise time.
h = 0.5g*Tf^2 = 703 m.
4.9Tf^2 = 703, Tf = Fall time = ?.
Tr+Tf = Time in light.
c. Range = X1*(Tr+Tf).
To find the maximum altitude, total time of flight, and horizontal range of the object, we can use the kinematic equations of motion. Let's break down the problem into these three parts:
(a) Maximum altitude reached by the object:
To determine the maximum altitude, we need to find the vertical component of the initial velocity at the highest point of the trajectory. We can use the equation:
Vf^2 = Vi^2 + 2 * a * d
where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and d is the displacement.
At the highest point, the final velocity is zero (since the vertical component is momentarily at rest). The initial vertical velocity (Vi) can be calculated as:
Vi = V * sin(theta)
where V is the initial speed and theta is the launch angle in radians.
We can calculate the displacement (d) using the equation:
d = (Vf^2 - Vi^2) / (2 * a)
Since Vf is zero, the equation simplifies to:
d = -Vi^2 / (2 * a)
The acceleration (a) is equal to the acceleration due to gravity, which is approximately -9.8 m/s^2 (considering upward as positive).
Now, we can substitute the values into the equation:
d = -Vi^2 / (2 * a)
Substituting the values, we have:
d = -V^2 * sin^2(theta) / (2 * a)
Calculating the value, we get:
d = -120^2 * sin^2(78º) / (2 * -9.8)
The negative sign indicates that we consider upward motion. Evaluating this expression, we find:
d ≈ 593.07 m
So, the maximum altitude reached by the object is approximately 593.07 meters.
(b) Total time of flight:
The total time of flight is the total time taken by the object from launch to landing. We can calculate it using the equation:
Time = (2 * Vi * sin(theta)) / a
Substituting the known values:
Time = (2 * 120 * sin(78º)) / -9.8
Evaluating this expression, we find:
Time ≈ 24.28 s
So, the total time of flight is approximately 24.28 seconds.
(c) Horizontal range:
The horizontal range is the horizontal distance covered by the object during its flight. We can calculate it using the equation:
Range = Vi * cos(theta) * Time
Substituting the values:
Range = 120 * cos(78º) * 24.28
Calculating the expression, we find:
Range ≈ 1347.9 m
So, the horizontal range of the object is approximately 1347.9 meters.
To summarize:
(a) The maximum altitude reached by the object is approximately 593.07 meters.
(b) The total time of flight is approximately 24.28 seconds.
(c) The horizontal range of the object is approximately 1347.9 meters.