A body is projected at an angle of 30 degree with the horizontal at initial speed of 200m/s. In how many seconds will it reach the ground? How far from the point of projection will it strike?
a. Vo = 200m/s[30o].
Xo = 200*Cos30 = 173.2 m/s.
Yo = 200*sin30 = 100 m/s.
Y = Yo + g*Tr = 0,
100 + (-9.8)Tr = 0,
Tr = 10.2 s. = Rise time.
Tf = Tr = 10.2 s. = Fall time.
Tr+Tf = 10.2 + 10.2 = 20.4 s. = Time to reach Gnd.
there is a "range equation"
wikipedia is a good source
You should have memorized two equations:
a. hf=hi+v*time+1/2 a t^2
where hf, hi are initiial velocities, v is the velocity in the direction of h, and a is the acceleration in the direction of height. In this case, it yields in upward motion..
hf=hi+muzzlevelocity*sinTheta*t-4.9t^2 and since hf, hi=0, you can find time of flight rather quickly.
b. distance equation: distance=velocity*time where velocity is in the direction of distance.
Here, range=v*cosTheta*time
And, if you look below, a number of the related equations are helpful.
b. d = Xo*(Tr + Tf) = 173.2 * 20.4
please help me solve this: A body is projected upward at an angle of 30° with the horizontal at an initial speed of 200m/s. In how many seconds will it reach the ground? How far from the point of projection will it strike?
To find the time it takes for the body to reach the ground, we can use the equation of motion for vertical motion:
s = ut + (1/2)gt^2
where:
s = displacement (which will be equal to zero when the body reaches the ground)
u = initial vertical velocity (which can be calculated by multiplying the initial speed by the sine of the angle)
t = time taken
g = acceleration due to gravity (approximately 9.8 m/s^2)
Since the body is projected at an angle of 30 degrees, the initial vertical velocity can be calculated as:
u = initial speed × sin(angle)
Given that the initial speed is 200 m/s and the angle is 30 degrees, we have:
u = 200 × sin(30)
u ≈ 100 m/s
Now, we can rearrange the equation to solve for time (t):
0 = ut + (1/2)gt^2
Substituting the values we know:
0 = 100t + (1/2)(9.8)t^2
Simplifying the equation:
0 = 100t + 4.9t^2
This is a quadratic equation. To solve, we can either factorize it or use the quadratic formula. Let's solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 4.9, b = 100, and c = 0. Solving:
t = (-100 ± √(100^2 - 4(4.9)(0)))/(2(4.9))
t = (-100 ± √(10000))/(9.8)
t ≈ (-100 ± 100)/9.8
Applying the quadratic formula:
t ≈ (-200/9.8) or t ≈ (0/9.8)
Since time cannot be negative, we discard the negative solution:
t ≈ 0/9.8
Therefore, the time it takes for the body to reach the ground is approximately 0 seconds.
To calculate the horizontal distance covered by the body, we can use the equation for horizontal motion:
s = ut
where:
s = horizontal distance covered
u = initial horizontal velocity (which can be calculated by multiplying the initial speed by the cosine of the angle)
t = time taken
Since the horizontal velocity remains constant throughout the projectile motion, the horizontal distance can be calculated by:
s = ut
Given that the initial speed is 200 m/s and the angle is 30 degrees, we have:
u = 200 × cos(30)
u ≈ 200 × √3/2
u ≈ 100√3 m/s
Since we have already determined that the time taken for the body to reach the ground is approximately 0 seconds, the horizontal distance covered is:
s = ut = (100√3 m/s) × (0 s) = 0 m
Therefore, the body will strike at a horizontal distance of 0 meters from the point of projection.