an object is released from an aeroplane which is diving at an angle of 30° from the horizontal with a speed of 50m/s. if the plane is at a height of 4000m from the ground when the object released, find a. the velocity of the object when it hits the ground. b) the time taken for the object to reach the ground.
Answers
Answer:
At any time t, a projectile's horizontal and vertical displacement are:
x = VtCos θ where V is the initial velocity, θ is the launch angle
y = VtSinθ – ½gt^2
The velocities are the time derivatives of displacement:
Vx = VCosθ (note that Vx does not depend on t, so Vx is constant)
Vy = VSinθ – gt
Velocity = Vxi + Vyj
The magnitude of velocity is √(Vx^2 + Vy^2)
At maximum height, Vy = 0 = VSinθ – gt
So at maximum height, t = (VSinθ)/g [total flight time = 2t]
The range R of a projectile launched at an angle θ with a velocity V is:
R = V^2 Sin2θ / g
The maximum height H is
H = V^2 Sin^2(θ) / 2g
In this case, V = 50m/s, θ = -30°, g = 9.81m/s^2
So Vx = VCosθ = 50(0.866) = 43.3m/s
Vy = VSinθ – gt = 50(-0.5) – 9.81t = -25 -9.81t
Total velocity = √[43.3^2 + (-25 – 9.81t)^2]
= √[2499.89 + 490.5t + 96.24t^2]
Unless t is known, this is as far as the solution can be taken.
Answer:
The velocity of the object when it hits the ground is √[2499.89 + 490.5t + 96.24t^2]
Explanation:
At any time t, a projectile's horizontal and vertical displacement are:
x = VtCos θ where V is the initial velocity, θ is the launch angle
y = VtSinθ – ½gt^2
The velocities are the time derivatives of displacement:
Vx = VCosθ (note that Vx does not depend on t, so Vx is constant)
Vy = VSinθ – gt
Velocity = Vxi + Vyj
The magnitude of velocity is √(Vx^2 + Vy^2)
At maximum height, Vy = 0 = VSinθ – gt
So at maximum height, t = (VSinθ)/g [total flight time = 2t]
The range R of a projectile launched at an angle θ with a velocity V is:
R = V^2 Sin2θ / g
The maximum height H is
H = V^2 Sin^2(θ) / 2g
In this case, V = 50m/s, θ = -30°, g = 9.81m/s^2
So Vx = VCosθ = 50(0.866) = 43.3m/s
Vy = VSinθ – gt = 50(-0.5) – 9.81t = -25 -9.81t
Total velocity = √[43.3^2 + (-25 – 9.81t)^2]
= √[2499.89 + 490.5t + 96.24t^2]
Unless t is known, this is as far as the solution can be taken.
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