A stone, 2kg is projected vertically at a velocity of 5m/s at an Angle of 60°
Calculate
I.Time taken to attain maximum height
II.maximum height reached
III.The range
Vo = 5m/s[60o].
Xo = 5*cos60 = 2.5 m/s.
Yo = 5*sin60 = 4.33 m/s.
1. Y = Yo + g*t.
0 = 4.33 - 9.8t, t = ?.
2. Y^2 = Yo^2 + 2g*h.
0 = 4.33^2 - 19.6h, h = ?.
3. Range = Vo^2*sin(2A)/g.
Range = 5^2*sin(120)/9.8 =
To calculate the time taken to attain maximum height, maximum height reached, and the range, we can use the following equations of motion:
For vertical motion:
1. Time taken to reach maximum height (T): T = (u sinθ) / g
For projectile motion:
2. Maximum height reached (H): H = (u^2 sin^2θ) / (2g)
3. Range (R): R = (u^2 sin2θ) / g
Where:
u = initial velocity (5 m/s)
θ = angle of projection (60°)
g = acceleration due to gravity (9.8 m/s^2)
Let's calculate each value step by step.
I. Time taken to attain maximum height (T):
T = (u sinθ) / g
= (5 * sin(60°)) / 9.8
≈ 0.85 seconds
II. Maximum height reached (H):
H = (u^2 sin^2θ) / (2g)
= (5^2 * sin^2(60°)) / (2 * 9.8)
≈ 0.64 meters
III. The range (R):
R = (u^2 sin2θ) / g
= (5^2 * sin(2*60°)) / 9.8
≈ 2.55 meters
Therefore,
I. Time taken to attain maximum height ≈ 0.85 seconds
II. Maximum height reached ≈ 0.64 meters
III. The range ≈ 2.55 meters
To calculate the time taken to attain maximum height, maximum height reached, and the range of a projectile, we can use the equations of projectile motion. These equations are derived using the principles of kinematics.
I. Time taken to attain maximum height:
The vertical motion of the projectile can be analyzed independently of the horizontal motion. We can use the equation for vertical displacement:
h = (v₀y²) / (2g)
Where:
h = maximum height reached
v₀y = initial vertical velocity
g = acceleration due to gravity (approximately 9.8 m/s²)
To find the initial vertical velocity (v₀y), we need to decompose the initial velocity into horizontal and vertical components. The vertical component can be determined using trigonometry:
v₀y = v₀ * sinθ
Where:
v₀ = initial velocity of the projectile
θ = angle of projection
Substituting this value into the equation for vertical displacement, we get:
h = [(v₀ * sinθ)²] / (2g)
Now, we can plug in the given values:
v₀ = 5 m/s
θ = 60° (converted to radians, which is π/3)
h = [(5 * sin(π/3))²] / (2 * 9.8)
Simplifying the equation, we find:
h ≈ 0.637 m
Now, to find the time taken to attain maximum height, we can use the equation for vertical motion:
v = v₀y - gt
Where:
v = final vertical velocity (at maximum height)
v₀y = initial vertical velocity
g = acceleration due to gravity
t = time taken
At maximum height, the final vertical velocity is zero (v = 0). Rearranging the equation, we can solve for time (t):
t = v₀y / g
Substituting the values:
v₀y = 5 * sin(π/3)
g = 9.8 m/s²
t ≈ (5 * sin(π/3)) / 9.8
Simplifying the equation, we find:
t ≈ 0.279 s
Therefore, the time taken to attain maximum height is approximately 0.279 seconds.
II. Maximum height reached:
We have already calculated the maximum height reached (h) earlier. The maximum height is approximately 0.637 meters.
III. The range:
The range of a projectile is the horizontal distance covered by the object before it returns to the ground. To find the range (R), we can use the equation for horizontal motion:
R = v₀x * 2t
Where:
R = range
v₀x = initial horizontal velocity
t = total time of flight
The initial horizontal velocity (v₀x) can be determined using trigonometry:
v₀x = v₀ * cosθ
Substituting it into the equation for range, we get:
R = v₀ * cosθ * 2t
We have already calculated the value of t as 0.279 seconds. Now, let's calculate v₀x:
v₀x = v₀ * cosθ
v₀ = 5 m/s
θ = 60° (converted to radians, which is π/3)
v₀x = 5 * cos(π/3)
Simplifying the equation, we find:
v₀x ≈ 2.5 m/s
Now, we can calculate the range:
R = 2.5 * 2 * 0.279
Simplifying the equation, we find:
R ≈ 1.39 m
Therefore, the range of the projectile is approximately 1.39 meters.