Answer :
Answer:
a) v₀ = 32.64 m / s , b) x = 59.68 m
Explanation:
a) This is a projectile launching exercise, we the distance and height of the cliff
x = v₀ₓ t
y = [tex]v_{oy}[/tex] t - ½ g t²
We look for the components of speed with trigonometry
sin 43 = v_{oy} / v₀
cos 43 = v₀ₓ / v₀
v_{oy} = v₀ sin 43
v₀ₓ = v₀ cos 43
Let's look for time in the first equation and substitute in the second
t = x / v₀ cos 43
y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²
y = x tan 43 - ½ g x² / v₀² cos² 43
1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²
v₀² = g x² / [(x tan 43 –y) 2 cos² 43]
Let's calculate
v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]
v₀ = √ (35280 / 33.11)
v₀ = 32.64 m / s
.b) we use the vertical distance equation with the speed found
y = [tex]v_{oy}[/tex] t - ½ g t²
.y = v₀ sin43 t - ½ g t²
25 = 32.64 sin 43 t - ½ 9.8 t²
4.9 t² - 22.26 t + 25 = 0
t² - 4.54 t + 5.10 = 0
We solve the second degree equation
t = (4.54 ±√(4.54 2 - 4 5.1)) / 2
t = (4.54 ± 0.46) / 2
t₁ = 2.50 s
t₂ = 2.04 s
The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled
x = v₀ₓ t
x = v₀ cos 43 t
x = 32.64 cos 43 2.50
x = 59.68 m
The minimum muzzle velocity be for the shell to clear the top of the cliff must be 56m/s and the projectile will continue to move on the ground at the top of the cliff if there is no friction.
The question represents the case of projectile motion.
Let y be the height of the cliff and x be the distance of the cannon from the cliff. Then,
y = 25m, and
x = 60m
Now the projectile is thrown at an angle of 43° with the horizontal.
Let its initial velocity be v.
The horizontal component of velocity, [tex]v_x = vcos(43)[/tex]
The vertical component of velocity, [tex]v_y = vsin(43)[/tex]
(a) Since there is no external force acting along horizontal direction, the equation of motion can be written as:
[tex]x = vcos(43)t\\ \\ t=\frac{x}{vcos(43)} [/tex]....(1)
In the vertical direction, gravitation force is acting, so the equation of motion:
[tex]y =vsin(43)*t+\frac{1}{2}g*t^2 [/tex]
substituting value of t from equation (1):
[tex]y=vsin(43)*\frac{x}{vcos(43)}-\frac{1}{2}g\frac{x^2}{v^2cos^2(43)} \\\\ v^2 = \frac{1}{2}\frac{gx^2}{[xtan(43)-y]cos^2(43)}\\ \\ v^2=\frac{1}{2}\frac{9.8*60^2}{[60*0.93-25]*0.73}\\ \\ v^2 = 3138 m^2/s^2\\ \\ v=56m/s[/tex]
Hence, the initial velocity must be 56m/s
(b) Since the ground at the top of the cliff is level at an elevation of 25 m, to calculate the distance of the projectile we need the value of the coefficient of friction to calculate the frictional force. In the absence of frictional force, the projectile will not stop since there is no horizontal force acting on it.
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