A bullet is fired from the ground at an angle of 60^o above the horizontal. What initial speed π£π£0 must the bullet have in order to hit a point 500 ft high on a tower located 800 ft away (ignoring air resistance) ?
Heavens, we are using feet ?
I guess the g = 32 ft/s^2 approximately if I remember from before the attack on Pearl Harbor.
Divide this into a vertical problem and a horizontal problem
I am going to call initial speed = s
Vertical
initial speed up = Vi = s sin 60 = .866 s
v = .866 s - 32 t
h = 0 + Vi t - 16 t^2
500 = .866 s t - 16 t^2
16 t^2 - .866 s t + 500 = 0
Horizontal
u = S cos 60 forever = 0.5 s
d = u t = 0.5 s t = 800
s t = 1600
t = 1600/s
---------------------------------now plug and chug
16 (1600^2/s^2) - .866 s(1600/s) + 500 = 0
16 (1600^2/s^2) - 1386 + 500 = 0
16 (1600^2/s^2) = 886
s^2 = 46230
s = 215 ft/second
Always check my arithmetic carefully :)
To find the initial speed π£π£β of the bullet, we can break down the horizontal and vertical components of its motion.
1. Vertical motion:
The initial height of the bullet is 0 ft (since it was fired from the ground), and it needs to reach a height of 500 ft. We can use the kinematic equation for vertical motion:
Ξπ¦ = π£βπ¦ * π‘ + 0.5π * π‘Β²
where Ξπ¦ is the vertical displacement (500 ft), π£βπ¦ is the initial vertical velocity, π‘ is the time of flight, and π is the acceleration due to gravity (-32.2 ft/sΒ²).
2. Horizontal motion:
The horizontal distance traveled by the bullet is 800 ft. We can again use the kinematic equation, but since there is no acceleration in the horizontal direction, the equation simplifies to:
Ξπ₯ = π£βπ₯ * π‘
where Ξπ₯ is the horizontal displacement (800 ft) and π£βπ₯ is the initial horizontal velocity.
3. Relationship between vertical and horizontal velocities:
The angle of 60Β° above the horizontal allows us to determine the relationship between the vertical and horizontal velocities. Since the angle is given, we know that:
tan(60Β°) = π£βπ¦/π£βπ₯
Solving for π£βπ¦, we get π£βπ¦ = tan(60Β°) * π£βπ₯.
4. Time of flight:
The time of flight can be determined from the vertical motion equation:
500 = π£βπ¦ * π‘ + 0.5 * (-32.2) * π‘Β²
Simplifying and rearranging the equation, we get:
16.1π‘Β² - π£βπ¦π‘ - 500 = 0
Apply the quadratic formula to solve for π‘, considering only the positive root.
5. Substitute π‘ into the horizontal motion equation:
Now that we have π‘, we can substitute it into the horizontal motion equation:
800 = π£βπ₯ * π‘
6. Substitute π£βπ¦ into the equation:
Using the relationship between π£βπ¦ and π£βπ₯, we can substitute π£βπ¦ into the equation:
800 = (π‘an(60Β°)) * π£βπ₯
7. Combine equations and solve for π£βπ₯:
Substituting the expression for π£βπ¦ from step 6 into the equation from step 5, we get:
800 = (tan(60Β°) * π‘) * π£βπ₯
Combining this equation with the equation from step 4, we can solve for π£βπ₯.
Solving these equations will give us the initial speed π£β of the bullet required to hit the point 500 ft high on the tower located 800 ft away.