Question

A golfer hits his tee shot along a flat fairway at 40 degrees to the horizontal with an initial speed of 50m/s. What is the range of the ball? What is the maximum height to which the ball rises?


Please answer clearly and explain the solution clearly and easy to understand. Thanks!
(guaranteed medal for a good answer)

Answer 1

\[Max.\:\:Height = \frac{u^2sin^2\theta}{2g}\]\[Max.\:\:Range = \frac{u^2sin2\theta}{g}\]
Use these and you will get you answer \(u\) is initial velocity and \(\theta\) is angle i.e 40.

Answer 2

i s the max height - ?

Answer 3

is the max height the same as
\[Max.\:\:Height = \frac{-(Vi \sin\theta)^2}{2g}\]
??

Answer 4

51.5 max height

Answer 5

hi