A golf ball is hit at a speed of 30 m/s at an angle of 450 above the horizontal. What is the maximum height of the ball?
(I get 22.96m, is this correct?)
2007-12-10 06:54:04 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Sorry, thanks for those who noticed that I really meant 45 above, not 450... Sorry to the others...
2007-12-10 07:18:43 · update #1
If that is 45degrees from the horizontal, then the initial vertical velocity is 30sin(45) = 15sqrt2
v(1)^2 - v(0)^2 = 2ax
x = -v(0)^2/2*a
a = -9.8
so x = -(15sqrt2)^2/2*-9.8 = 22.96m, correct.
2007-12-10 07:08:29 · answer #1 · answered by eazylee369 4 · 0⤊ 0⤋
vertical component of velocity = 30 cos 45 deg = (1/2)30 sqrt 2
the kinetic energy of the vertical component of velocity = mgh = (1/2) m v^2 = (1/2) m (1/4) (30)(30) (2) = 900/4 X M
h = (900/4 }(1/g) = 225/9.8= 22.96 m OK
Of course this ignore the frictional effects of the motion through the air and, more importantly, ignores the aerodynamic effects. The fact that the ball is dimpled rather than smooth allows it to go higher and farther.
2007-12-10 15:06:54 · answer #2 · answered by LucaPacioli1492 7 · 0⤊ 0⤋
450 =360+90
i.e 90 degrees thats straight up so.........
v^2=u^2+2*g*h
0=900-2*9.8*h
h=45.918 metres
2007-12-10 15:05:05 · answer #3 · answered by shivam 2 · 0⤊ 1⤋