Question

A golfer hits a golf ball at an angle of 25.0° to the ground. If the golf ball
covers a horizontal distance of 301.5 m, what is the ball’s maximum
height? (Hint: At the top of its flight, the ball’s vertical velocity component will be zero.

Answer 1

are you sure, you're not missing any data?

Answer 2

This can be solved using the equations of projectile motion with constant acceleration. Since, there is never any acceleration in the horizontal direction, the horizontal displacement can be expressed as: \[\Delta x = v_{x}t\]where v_x is the horizontal component of the velocity, and t is the flight time.
Since gravity is the only force acting on the projectile and gravity only acts in the vertical component, the height can be expressed as:\[\Delta y = v_yt - {1 \over 2} g t^2\]The projectile follows a parabolic path, therefore the time at which the maximum height occurs can be found by solving the derivative of the equation for y for zero. This is expressed as:\[0 = v_y - g t \rightarrow t = {v_y \over g}\]The time is takes for the projectile to reach the ground is twice the time it takes the projectile to reach its maximum height, therefore, \[t = {2 v_y \over g}\]We still don't know the time the projectile is in flight, but we do know the maximum range. The range of the projectile can be found by solving the equation for x by substituting the expression for the time it takes the projectile to reach the ground. This can be expressed as:\[r = v_x {2v_y \over g}\]From trigonometry, \[\begin{matrix} v_x = \cos(\theta)\\ v_y = \sin(\theta)\end{matrix}\]Substituting these into the equation for the range, we get \[r = {2\cos(\theta)\sin(\theta) \over g}\]We know from our trigonometric identities that \[\sin(2\theta) = 2\sin(\theta)\cos(\theta)\]Therefore, \[r = {v^2 \over g} \sin(2\theta)\]We know the range, the angle theta, and gravity; therefore, we can solve the range equation for velocity.
From here we can go two different directions, we can solve for the horizontal component of the velocity and use this to find the time it takes to reach the maximum height. Then plug this time into the equation for y. A second option is to substitute the expression for time at the maximum height into the equation for y. This can be expressed as:\[\ y_{max} = {v_y^2 \over g} - {1 \over 2}g \left( v_y \over g \right)^2 = {v_y^2 \over g} - {1 \over 2}{v_y^2 \over g} = {v^2 \sin^2(\theta) \over 2g}\]We can plug in the velocity we found earlier to find the maximum height.

Answer 3

you can find of y max when x is zero.

Answer 4

Correction. "From here we can go two different directions, we can solve for the horizontal component..." Replace horizontal with vertical.

Answer 5

that is very detailed. thanks a lot! :)

Answer 6

your way of solving is no doubt a very good way. but i am skeptical though, i don't think the book has mentioned the need to use trigonometric identities and derivatives. So I am trying to find a simpler way, perhaps, to solve the problem. The only thing is that, like esteban said, the question seems to miss some pieces of data.

Answer 7

R=301.5
\[\theta \]=25
g=10 \[m/s^2\]
V(0)=?
Y(max)=?
we know that the range van be measure by the following equation:
R=\[R=v^2*\sin (2\theta)/g\]
so by putting the numbers in the equation v ll find out that V^2=3935m/s
we also know that at the highest point V(y)=0
so we v can write down:
\[V(y)^2-V(0)(y)^2=-2g \Delta y\]
we also know that \[V(y)=V(0)*\sin \theta\]
now the following equation would be:
\[V(y)^2-[V(0)\sin \theta]^2=-2g \Delta y\]
so the y which is the highest point would be:35m

Answer 8

anhhuyalex. Without knowing which particular physics course you are taking, I wanted to be as thorough as possible. Have faith that the equations I have presented you with are correct. If you not in a calculus based physics class, your book should give you the equations of projectile motion. These are essential equations to basic physics.

Although your book does not mention the use of derivatives or trigonometric identities, they are used to derive the equation of projectile motion.

Let me present you with some key equations relating to projectile motion that you will need to use to solve this problem. \[\begin{matrix} \text{Horizontal Velocity} & v_x = v \cos(\theta) & (1)\\ \text{Vertical Velocity} & v_y = v \sin(\theta) & (2)\\ \text{Time \to reach maximum height} & t_{y_\max} = {v_y \over g} & (3) \\ \text{ Time \to reach ground} & t_{ground} = {2v_y \over g} & (4)\\ \text{Horizontal Position} & x(t) = v_x t & (5)\\ \text{Vertical Position} & y(t) = v_yt - {1 \over 2} g t^2 & (6)\\ \text{Range of projectile} & r = x_{\max} = {v^2 \over 2} \sin(2 \theta) & (7)\\ \text{Maximum Height} & y_{\max} = {v^2 \sin^2(\theta) \over 2g} & (8)\end{matrix}\]We know the range and theta. Therefore, we can use Equation (7) to find the velocity of the projectile. Then we can use Equation (8) to find the maximum height of the projectile.

Sorry if I caused any confusion.