Dynamics Question

BASHIR

A ball is struck such that its max height is equal to its range. determine the angle its struck at. Air resistance is ignored and the ground is horizontal.

my reasoning is that:

using vectors
Xsin(angle) = Xcos(angle)
so tan(angle) = x/x
so inverse tan of 1 is 45degrees

Would this be the right approach to the question
Id ask my lecturer but im at home
Cheers

Overature

yeah it is. think about it as a square where the height = range = side of square. the angle would have to be 45 if this was to be true. any other angle and the height would not equal the range

BASHIR

Hey just found out
It seems I was a little wrong
Range =v^2sin(2θ)/g
Max height = (vsinθ)^2/2g

Range =v2sin(2θ)/g=(vsinθ)2/2g= Max height

4sinθcosθ=sinθsinθ because sin2θ=2sinθcosθ
4cosθ=sinθ
tanθ=4

θ=tan(inverse)4
= 75.96 degrees
Problem is Xcosθ=Xsinθ equates horizontal and vertical components of X (whatever X is), which isn't same as equating range of projectile to maximum possible height when thrown at certain angle

Cheers for the reply anyways

breadmonkey

Dynamics?

BASHIR

Yes dynamics, projectile motion

Dunphus

BASHIR wrote: »

Yes dynamics, projectile motion

I was confused by this at first. You're right of course, this is dynamics but I studied this as mechanics. The dynamics I studied, and breadmonkey too I'm sure, would be time dependent, wind, waves, earthquakes, differential equations, equations of motion etc.

But don't worry, it's all ahead of you

BASHIR

Ha I hope not Doing finals in 4th year mechanical!!!
Ya you would think that question would be easy, just copped it myself, duh kinda moment.