- #1
robcowlam
- 9
- 0
I am trying to calculate the initial velocity, u and angle of projection(theta) of a projectile. The projectile is fired from 1.06m above the ground and from this point reaches a maximum of 2.12m from the ground. The distance traveled in the x-direction when the projectile reaches its maximum height is 1m.
This is all the information given in the question.
I am taking my x-axis to be the horizontal direction and the y-axis the vertical direction, calling the point where the projectile is fired from the origin.
I have tried to get a solution in the following way:
Motion in y-direction:
a=dv/dt
a dt = dv
integrating both sides gives:
a.t = 0 - uSin(theta) (where u is initial velocity and v is final velocity)
dividing by a gives an expression for the time taken to reach maximum height.
Motion in the x-direction:
S=ut +0.5at^2
however a = 0
so; 1=uCos(theta).t
Substituting in t from earlier gives:
9.81=(u^2) Cos(theta)Sin(theta) from which i get u^2=9.81/sin(theta)cos(theta)
In the y direction again:
v^2 -u^2 = 2as
20.8=9.81/sin(theta)cos(theta)
9.81/20.8=sin(theta)cos(theta)
2(0.47)=2sin(theta)cos(theta) using the trig ID sin(2A)=SinACosA
sin(2theta)=0.94
and theta =35degrees;
Any help would be GREATLY appreciated
Thanks!
However this is not the answer given with the problem
This is all the information given in the question.
I am taking my x-axis to be the horizontal direction and the y-axis the vertical direction, calling the point where the projectile is fired from the origin.
I have tried to get a solution in the following way:
Motion in y-direction:
a=dv/dt
a dt = dv
integrating both sides gives:
a.t = 0 - uSin(theta) (where u is initial velocity and v is final velocity)
dividing by a gives an expression for the time taken to reach maximum height.
Motion in the x-direction:
S=ut +0.5at^2
however a = 0
so; 1=uCos(theta).t
Substituting in t from earlier gives:
9.81=(u^2) Cos(theta)Sin(theta) from which i get u^2=9.81/sin(theta)cos(theta)
In the y direction again:
v^2 -u^2 = 2as
20.8=9.81/sin(theta)cos(theta)
9.81/20.8=sin(theta)cos(theta)
2(0.47)=2sin(theta)cos(theta) using the trig ID sin(2A)=SinACosA
sin(2theta)=0.94
and theta =35degrees;
Any help would be GREATLY appreciated
Thanks!
However this is not the answer given with the problem