A projectile is shot from the edge of a cliff 145m above ground level with an initial speed of 65.0 m/s at an angle of 37.0° with the horizontal. How far did the shot go horizontally?
Also find the final velocity in y (v_fy) of the shot just before it hits the ground.
horizontal components
Vox=65cos(37)=51.11 = Vfx
verticall components
Voy=65sin(37)=39.11 ==> Vfy=-39.11
find t from this formula Vfy=Voy-gt and solve for t. t=7.98s
plug t in this equation X=Voxt+0.5(a)t^2
X=414.32
Range = Vo^2*sin(2A)/g.
Range = 65^2*sin(2*37)/9.8 = 414.4 m.
To find the horizontal distance traveled by the projectile, we need to calculate the horizontal component of its initial velocity.
The horizontal component of the initial velocity can be found using the formula:
Vx = V * cos(theta)
where:
Vx is the horizontal component of velocity,
V is the initial velocity of the projectile,
and theta is the angle between the initial velocity vector and the horizontal axis.
In this case:
V = 65.0 m/s (given)
theta = 37° (given)
Using these values in the formula, we can calculate the horizontal component of the initial velocity:
Vx = 65.0 m/s * cos(37°)
Now, let's find the value of cos(37°) using a calculator:
cos(37°) ≈ 0.7986
Now we can substitute this value into the formula:
Vx = 65.0 m/s * 0.7986
Vx ≈ 51.91 m/s
Therefore, the horizontal component of the initial velocity is approximately 51.91 m/s.
To calculate the horizontal distance traveled by the projectile, we can use the equation of motion for horizontal motion:
Distance = Velocity * Time
Since we are interested in the distance traveled horizontally, and there are no horizontal forces acting on the projectile (ignoring air resistance), the horizontal component of velocity remains constant throughout the motion.
We need to find the time of flight for the projectile. The time of flight is the total time it takes for the projectile to reach the ground. We can use the equation for vertical motion to find this time:
y = y0 + Vyt - 1/2 * g * t^2
where:
y is the vertical displacement from the starting point,
y0 is the initial vertical position,
Vy is the vertical component of velocity,
g is the acceleration due to gravity (approximately -9.8 m/s^2),
and t is the time.
In this case:
y0 = 145 m (given)
y = 0 m (since it reaches the ground)
Vy = V * sin(theta)
g = -9.8 m/s^2
Using these values, we can rearrange the equation to solve for t:
y = y0 + Vy * t - 1/2 * g * t^2
0 = 145 m + V * sin(theta) * t - 1/2 * (-9.8 m/s^2) * t^2
Simplifying the equation:
4.9 m/s^2 * t^2 - V * sin(theta) * t - 145 m = 0
This is a quadratic equation in terms of time. We can solve for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where:
a = 4.9 m/s^2
b = -V * sin(theta)
c = -145 m
Using these values, we can calculate the time of flight:
t = [ -(-V * sin(theta)) ± √((-V * sin(theta))^2 - 4 * (4.9 m/s^2) * (-145 m)) ] / (2 * 4.9 m/s^2)
Substituting the known values:
t = [ V * sin(theta) ± √((V * sin(theta))^2 + 4 * 4.9 m/s^2 * 145 m) ] / (9.8 m/s^2)
Now let's substitute the known values and calculate the time of flight:
t = [65.0 m/s * sin(37°) ± √((65.0 m/s * sin(37°))^2 + 4 * 4.9 m/s^2 * 145 m)] / (9.8 m/s^2)
Using a calculator, we find:
t ≈ 9.18 s
Therefore, the time of flight of the projectile is approximately 9.18 seconds.
Now that we know the horizontal component of velocity (Vx) and the time of flight (t), we can calculate the horizontal distance traveled by the projectile:
Distance = Vx * t
Substituting the known values:
Distance = 51.91 m/s * 9.18 s
Calculating the value:
Distance ≈ 476.65 m
Therefore, the projectile traveled approximately 476.65 meters horizontally.