Robb
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Homework Statement
Homework Equations
kinematics.
a=F/m
The Attempt at a Solution
100cos(70)=342
100sin(70)=393.7
X=1/2(-9.8)t^2+100t; t=0. t=20.4
The discussion revolves around a projectile motion problem involving the calculation of height, range, and time of flight. The context includes kinematic equations and the effects of initial velocity components derived from a launch angle.
Some participants have provided calculations for initial velocities and questioned the validity of their results, particularly regarding time of flight. Others are exploring the implications of the building's height on the projectile's motion and whether a symbolic answer is appropriate. The discussion is active with various interpretations being considered.
The problem lacks specific information about the height of the building, leading to uncertainty in how to approach the range and time of flight calculations. Participants are debating whether to assume a height or consider the launch and landing elevations to be the same.
You might want to check the arithmetic above before going further in this calculation. Also, don't forget to indicate units.Robb said:Homework Statement
See attached.
Homework Equations
Kinematic equations.
The Attempt at a Solution
100cos(70)=342 (x component)
100sin(70)=393.7 (y component)
X=1/2(-9.8)t^2 + 100t = -4.9t^2 + 100t; t=0, t=20.4
V(final)= V(initial) + at =100 + (-9.8)(20.4)=-99.9m/s
I'm convinced t is not equal to 20.4.
The g in this case is just the constant g without sign. You might try deriving the expression for the maximum height to see why this is.Robb said:h=[v(initial)^2*sin(theta)^2]/2g = [(100)^2(sin70)^2]/-19.6 = -450.5
Not sure how this can be negative unless it is accounting for the height of the building?