It is observed that an object that is thrown with projectile motion on a horizontal plane and reaches a maximum height of 38.5m and travels horizontally a distance equivalent to four times the maximum height. Determine the launch velocity (in m/s).
vertical problem:
v = Vi - g t
at top v = 0
g t = Vi
h = 0 + Vi t - (g/2) t^2
38.5 = g t^2 - (1/2) g t^2 = 4.9 t^2 on earth
t^2 = 38.5/4.9
t = 2.8 seconds upward (so 5.6 seconds in the air)
Vi = 9.8 t = 9.8 * 2.8 = 27.4 m/s
==================
Now horizontal problem x = 4 *38.5 = 154 meters
time = 5.6 seconds
so
Vx = 154/5.6 = 27.5 m/s
LOL --> launch at about 45 degrees :) v = about 1.41 * 27.45
v^2 = 27.5^2 + 27.4*2
yes, the 45° angle is expected. Since we know the vertex of the parabola is at (2h,h) and y=0 at 0 and 4h,
y = h - 1/(4h) (x-2h)^2
y' = -1/(2h) (x-2h)
tanθ = y'(0) = 1
Now you have another handy fact to recall, which could save you some calculation in the future.
To determine the launch velocity of the object, we can break down the problem into two components: vertical motion and horizontal motion.
Vertical Motion:
When the object reaches its maximum height, its vertical velocity component becomes zero. We can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2ad
Where:
vf = final vertical velocity (0 m/s at the maximum height)
vi = initial vertical velocity (launch velocity)
a = vertical acceleration (acceleration due to gravity, -9.8 m/s^2)
d = vertical displacement (38.5 m)
Rearranging the equation, we have:
0 = vi^2 + 2(-9.8)(38.5)
Simplifying further:
0 = vi^2 - 763.6
vi^2 = 763.6
Taking the square root of both sides to solve for vi:
vi = √763.6
Horizontal Motion:
Given that the object travels horizontally a distance equivalent to four times the maximum height, we can determine the horizontal displacement (d) using the formula:
d = vix * t
Where:
d = horizontal displacement (4 times the maximum height, 4 * 38.5 = 154 m)
vix = initial horizontal velocity (same as the launch velocity)
t = time of flight
To find the time of flight, we can use the vertical motion equation:
vf = vi + at
Where:
vf = final vertical velocity (0 m/s at the maximum height)
vi = initial vertical velocity (launch velocity)
a = vertical acceleration (acceleration due to gravity, -9.8 m/s^2)
t = time of flight
Rearranging the equation, we have:
0 = vi + (-9.8)t
vi = 9.8t
Substituting this expression for vi in the horizontal motion equation:
d = (9.8t) * t
Simplifying further:
d = 9.8t^2
Now, we can equate the horizontal displacement to 154 m:
154 = 9.8t^2
Dividing both sides by 9.8:
t^2 = 15.7143
Taking the square root of both sides to solve for t:
t = √15.7143
Solving for vi:
We can solve for vi by substituting the value of t in the equation vi = 9.8t:
vi = 9.8 * √15.7143
Calculating the value of vi using a calculator or software:
vi ≈ 14.069 m/s
Therefore, the launch velocity of the object is approximately 14.069 m/s.