can you help me with a physics problem?

Question

A golfer can hit a golf ball a horizontal distance of over 300 m on a good drive. What maximum height will a 301.5 m drive reach if it is launched at an angle of 25 degree to the ground? (Hint: at the top of its flight, the ball's vertical velocity component will be zero.)

Please help me solve this problem with calculations/explainations. Thank you

Answer 1

Its a pain to do this with X and Y components, but there is an easier way....

Range Formula - used when start and end heights are the same
Q = Theta
d = ((Vo^2)(sin 2Q))/g

301.5 = ((Vo^2)(sin 2(25)))/9.81

Vo = 62.13 m/s

Now that u have initial velocity, find the y component

Vo Sin Q

62.13 sin 25 = 26.26 m/s

Then No Time Formula

Vf^2 = Vo^2 + 2gd

0 = (26.26^2) + 2(-9.81)d

d = 35.15 m

Hope this helps

Answer 2

Use distance = 1/2 acceleration x time squared. Anything thrown or hit like a golf ball follows a parabolic path, where the first half is a mirror image of the second half in both distance and speed (so idiots who fire bullets into the air sometimes kill people because bullets come down just as fast as they go up).

You need trigonometry to figure out how the angle affects how high the ball goes. You already know that the vertical velocity changes but the horizontal velocity stays constant, more or less.

Answer 3

When the ball launched it has both vertical (vy) and horizontal (vx) velocity components. If the launch is with velocity v0 at an angle ø with the ground, the vertical component vy0 is v*sin(ø) and the horizontal component vx0 is v*cos(ø). These are considered independently. The vertical velocity will be vy0 - g*t, where g=accel due to gravity = 9.8m/sec^2. As you said, maximum height is when vertical velocity gets to zero. From the formula vy0 - g*t = 0, compute t. This is the time it takes to reach max height. The formula for height of an object is h = vy0*t - .5*g*t^2. Use the t just computed find the max height.

Answer 4

firstly i broke the velocity into its vectors where Vx(velocity in the x direction)=Vcos25 and Vy=Vsin25
Now i made equations that applied out of the kinematic equations
Vx*(multiply)t(time)=301.5m that makes sense, your speed times the time gives you the distance.
Now using the equation V=Vo + at; we apply the Y velocity to make the equation 0=Vy -9.8*1/2t we had to use 1/2t because at half the total time we will be at our max height.
from here you should be able to figure it out with some hard calculation. first you need to substitute Vx with Vcos25 and same with Vy. Now you have 2 equations with only V and t as variables. you need to solve for one then plug it in to get the other value.
you should have the initial value of V. now we can apply the equation V^2=Vo^2+2ady; where dy=change in y, or the exact number we want.