how far will a stone travel over level ground if it is thrown upward at an angle 30 with the horizontal and with a speed of 12m/s?what maximum range could be achieved with the same initial speed?
horizontal velocity=12cos30
vertical initial velocity:12sin30
time in air:
hf=hi+vi'*t-4.8t^2
hf=0 Hi=? assume zero, as stupid as that is.
0=12sin30*t-4.8t^2
t=sqrt(12sin30/4.8)
the horizontal distance:
horizontal velocity above*timeinair
Max range will occur at theta 45 deg. Figure it as above.
Love
To determine the distance a stone will travel over level ground when thrown upward at an angle of 30 degrees with the horizontal and a speed of 12 m/s, we need to calculate the horizontal component of the stone's initial velocity and the time it takes for the stone to reach its highest point.
1. Horizontal component of initial velocity (Vx):
Vx = V * cos(theta)
Vx = 12 m/s * cos(30 degrees)
Vx = 12 m/s * √3/2
Vx ≈ 10.39 m/s
2. Time to reach highest point (t_up):
Use the equation: t_up = Vy / g
where Vy is the vertical component of the initial velocity and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Vy = V * sin(theta)
Vy = 12 m/s * sin(30 degrees)
Vy = 12 m/s * 1/2
Vy = 6 m/s
t_up = 6 m/s / 9.8 m/s^2
t_up ≈ 0.61 s
3. Distance traveled horizontally at the peak (D_peak):
D_peak = Vx * t_up
D_peak = 10.39 m/s * 0.61 s
D_peak ≈ 6.35 m
4. Total distance traveled over level ground (D_total):
As the stone reaches the same height on its descent as it did on its ascent, the total distance traveled is twice the distance at the peak (D_peak).
D_total = 2 * D_peak
D_total = 2 * 6.35 m
D_total ≈ 12.7 m
Therefore, the stone will travel approximately 12.7 meters over level ground when thrown upward at an angle of 30 degrees with a speed of 12 m/s.
To find the maximum range that could be achieved with the same initial speed, we need to find the angle that provides the maximum horizontal distance.
5. Maximum range angle (theta_range):
The maximum range is achieved when the angle of projection (theta) is 45 degrees.
Therefore, the maximum range that could be achieved with the same initial speed of 12 m/s is when the stone is thrown at an angle of 45 degrees with the horizontal.
To find out how far a stone will travel over level ground when thrown upward at an angle of 30 degrees with the horizontal and a speed of 12 m/s, we can use the equations of projectile motion.
First, let's break the motion into horizontal and vertical components:
The initial velocity of the stone can be split into its horizontal and vertical components as follows:
Horizontal component: Vx = V * cos(theta)
Vertical component: Vy = V * sin(theta)
Where:
V is the magnitude of the initial velocity (12 m/s)
theta is the angle of projection (30 degrees)
Now, let's find the time it takes for the stone to reach the peak of its trajectory (when it stops moving upwards and starts falling down). At this point, the vertical component of velocity becomes zero.
Using the vertical motion equation:
Vy = Vy0 + a * t
Since the stone is thrown upwards, the initial vertical velocity (Vy0) is positive. At the peak, Vy becomes zero, and acceleration (a) due to gravity is -9.8 m/s^2 (assuming no air resistance).
0 = V * sin(theta) + (-9.8) * t_peak
Solving for t_peak, we get:
t_peak = (V * sin(theta)) / 9.8
Next, let's find the total time of flight, which is the time it takes for the stone to reach the ground again. Since the stone goes up to the peak, pauses momentarily, and then falls back to the ground, the total time of flight (T) is twice the time it took to reach the peak.
T = 2 * t_peak
Now, let's find the horizontal distance traveled by the stone. The horizontal distance (range) is given by:
Range = Vx * T
Substituting the values we have:
Range = (V * cos(theta)) * (2 * ((V * sin(theta)) / 9.8))
Simplifying further:
Range = (2 * V^2 * cos(theta) * sin(theta)) / 9.8
Now, let's calculate the range:
Range = (2 * (12^2) * cos(30) * sin(30)) / 9.8
Range ≈ 17.3 meters
So, the stone will travel approximately 17.3 meters over level ground when thrown upward at an angle of 30 degrees with a speed of 12 m/s.
To find the maximum range that could be achieved with the same initial speed, we need to find the angle of projection that maximizes the range. This angle, known as the optimal angle, can be calculated using calculus or by considering theoretical principles.
The maximum range is achieved when the angle of projection is 45 degrees (assuming no air resistance). This means that if the stone is thrown at 45 degrees with a speed of 12 m/s, it will travel the maximum distance.
Therefore, the maximum range that could be achieved with the same initial speed of 12 m/s is also approximately 17.3 meters.