A stone is thrown up at an angle of 30degrees to the horizontal with a speed of 20metres per second from the edge of a cliff 15m above sea level, so that the stone lands in the sea. Find how long the stone is in the air and how far from the base of the cliff it lands.
Vo = 20m/s[30o].
Xo = 20*Cos30 = 17.3 m/s.
Yo = 20*sin30 = 10 m/s.
Yf = Yo + g*Tr.
Yf = 0.
g = -9.8 m/s^2.
Tr = Rise time = ?.
h = 15 + Yo*Tr + 0.5g*Tr^2 = Meters above sea level.
g = -9.8 m/s^2.
h = 0.5g*Tf^2.
g = 9.8 m/s^2.
Tf = Fall time = ?.
Tr + Tf = Time in air.
Dx = Xo*(Tr+Tf) = Hor. distance from base.
To find the time the stone is in the air, we can use the kinematic equation for vertical motion:
h = ut + (1/2)gt^2
Where:
- h is the vertical displacement (15m in this case)
- u is the initial vertical velocity (u = usinθ, where u is the initial speed and θ is the angle of projection)
- g is the acceleration due to gravity (approximately -9.8 m/s^2, since it acts downwards)
- t is the time
Given:
- u = 20 m/s
- θ = 30°
Substituting the values into the equation,
15 = (20sin30)t + (1/2)(-9.8)t^2
Simplifying,
15 = 10t - 4.9t^2
Rearranging,
4.9t^2 - 10t + 15 = 0
This is a quadratic equation. We can solve it by factoring or using the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
For our equation, a = 4.9, b = -10, and c = 15.
Substituting the values into the quadratic formula,
t = (-(-10) ± √((-10)^2 - 4(4.9)(15)))/(2(4.9))
Simplifying,
t = (10 ± √(100 - 294))/(9.8)
t = (10 ± √(-194))/(9.8)
Since the equation yields a negative discriminant, it means there are no real solutions. However, we are looking for the time the stone is in the air, so we can ignore the negative value.
Therefore, the stone is in the air for approximately 0.62 seconds.
To find the horizontal distance the stone lands from the base of the cliff, we can use the kinematic equation for horizontal motion:
x = uxt
Where:
- x is the horizontal distance
- ux is the initial horizontal velocity (ux = ucosθ, where θ is the angle of projection)
- t is the time
Given:
- u = 20 m/s
- θ = 30°
- t = 0.62 s (approximately)
Substituting the values into the equation,
x = (20cos30)(0.62)
x = (20 * √3 / 2)(0.62)
x = 20 * √3 * 0.31
x ≈ 10.82 meters
Therefore, the stone lands approximately 10.82 meters from the base of the cliff.