A pistol is shot upwards at an angle of 37 degrees above horizontal, with an initial velocity of 375m/s. What height does it reach? What is its horizontal range?
(I pretty much know what formulas to use, but I'm not quite sure HOW to use them.. Also you should know that this is Freshman Physics so we don't use calculus.
Thank you)
no calculus needed. Just plug in your formulas:
h(x) = tanθ x - g/(2 (v cosθ)^2) x^2
= 0.75x - .0000547 x^2
That is just a parabola with vertex at
(6855.6,2570.8)
I expect you can find x when h=0.
To find the height reached by the pistol and its horizontal range, we can use the following formulas:
1. Vertical Displacement:
y = v₀y * t + (1/2) * a * t²
2. Horizontal Displacement:
x = v₀x * t
where:
- y is the vertical displacement or height reached by the pistol
- x is the horizontal displacement or range of the pistol
- v₀y is the initial vertical component of velocity (375 m/s * sin(37°))
- v₀x is the initial horizontal component of velocity (375 m/s * cos(37°))
- a is the acceleration due to gravity (-9.8 m/s²)
- t is the time of flight
Let's first calculate the time of flight using the vertical motion formula:
1. Determine the vertical component of the initial velocity:
v₀y = 375 m/s * sin(37°)
v₀y ≈ 225.455 m/s
2. Calculate the time of flight:
Use the formula: y = v₀y*t + (1/2)*a*t²
Since the pistol reaches its highest point when the vertical displacement (y) is maximum, we can set y = 0 and solve for t.
0 = (225.455 m/s) * t + (1/2) * (-9.8 m/s²) * t²
Rearranging and factoring out t, we get:
t * (225.455 m/s - (1/2) * 9.8 m/s² * t) = 0
So, t = 0 or t = (225.455 m/s)/(1/2 * 9.8 m/s²)
t = 0 or t ≈ 45.991 seconds
Since the pistol cannot stay in the air for negative time, we disregard t = 0.
3. Calculate the horizontal component of the initial velocity:
v₀x = 375 m/s * cos(37°)
v₀x ≈ 299.769 m/s
Now, we can find the height and horizontal range using the time of flight (t) we calculated.
4. Calculate the height reached (vertical displacement):
Using the vertical motion formula, substitute v₀y and t:
y = (225.455 m/s) * (45.991 s) + (1/2) * (-9.8 m/s²) * (45.991 s)²
y ≈ 10009.085 m
So, the pistol reaches a height of approximately 10009.085 meters.
5. Calculate the horizontal range (displacement):
Using the horizontal motion formula, substitute v₀x and t:
x = (299.769 m/s) * (45.991 s)
x ≈ 13775.052 m
Therefore, the horizontal range of the pistol is approximately 13775.052 meters.