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how to find flight time/distance and greatest hight of projectiles?

is acceleration, v is velocity, is position

vertically:

a = - g

v = u - gt (by integrating with respect to t and setting initial speed u as constant of integration)

s = ut -gt2/2 + s0 (by integrating again and setting initial position s0 as constant of integration, although object if usually projected from origin so s= 0)

horizontally:

a = 0

v = u (constant velocity as acceleration is zero)

s = ut + s0 (again s0 usually 0)

so for an object projected from the origin at speed u at an angle θ from the horizontal, the initial speed in the x direction is ucos(θ) and usin(θ) in the y direction

thus sx = utcos(θ) and sy = utsin(θ) - gt2/2

flight time: this is the value of t when sy returns to zero

0 = utsin(θ) - gt2/2

so either t = 0 (at launch) or usin(θ) - gt/2 = 0

=> flight time is t = 2usin(θ)/g

flight distance: this is the value of sx when sy = 0

when sy = 0, t = 2usin(θ)/g 

so sx = utcos(θ) = 2u2sin(θ)cos(θ)/g = u2sin(2θ)/g (as sin(2θ) = 2sin(θ)cos(θ))

=> flight distance is x = u2sin(2θ)/g

greatest hight:

this is when sy is at a stationary point, ie. when dsy/dt = 0. This is also when vy = 0.

vy = usin(θ) - gt = 0

=> t = usin(θ)/g

=> sy = utsin(θ) - gt2/2 = u2sin2(θ)/g -  u2sin2(θ)/2g =  u2sin2(θ)/2g

=> greatest height is h = u2sin2(θ)/2g


:)

Ben J. A Level Maths tutor, A Level Further Mathematics  tutor

2 years ago

Answered by Ben, an A Level Maths tutor with MyTutor


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