# Projectile motion of Golf ball- Can projection height be calculated just from initial velocity and travel time? [closed]

A Golf ball projected with a horizontal velocity of 30 meter per second and takes 4 second to reach the ground. Calculate the height the golf ball was projected from. Calculate the magnitude of golf ball's vertical velocity component just before hitting ground.

Is this problem solvable? Can I assume the projection angle was 45 degree?

I attempted as following:

Horizontal acceleration is zero so,

$$v_x(t) = v_x(0),$$

therefore the horizontal displacement is,

$$\Delta x = v_x(0) t = 30 \cdot 4 = 120\ m.$$

Vertical displacement is,

$$\Delta y = y(t) - y(0) = v(0) \sin\theta - \frac{gt^2}{2} = 30 \sin\theta - 0.5 \cdot 9.8 \cdot 4^2 = 30 \sin\theta - 78.4. \tag{1}$$

I suspect I am wrong in assuming $v(0) = 30\ m/s$ because the question only states the horizontal velocity.

If actually $v(0)$ is $30\ m/s$ then I can take it further as,

$$v(0) = v_x(0) \cos\theta$$

$$30 = 30 \cos\theta$$

$$\cos\theta = 1$$

$$\theta = 0^\circ$$

Then replacing $\theta$ in (1) gives $\Delta y = 78.4\ m$ which is definitely wrong.

## closed as off-topic by Jim, HDE 226868, Kyle Kanos, John Rennie, MartinAug 25 '15 at 12:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Jim, HDE 226868, Kyle Kanos, John Rennie, Martin
If this question can be reworded to fit the rules in the help center, please edit the question.

• Hint: no projection angle or vertical velocity component was specified. What does that tell you? – Gert Aug 24 '15 at 17:39
• damn, I typed my whole attempt and the edit time passed and I lost it. Let me type again.. I came up with an unreasonable value so surely followed a wrong path.. – user733105 Aug 24 '15 at 17:48
• @AcidJazz added my attempt – user733105 Aug 24 '15 at 18:04
• @Gert that tells me to assume angle was 0 degree so no vertical component was involved or given.. and ball was rolling? What else does it tell? :) – user733105 Aug 24 '15 at 18:06
• Stronger hint, read Gert's hint slowly, and this: physicsclassroom.com/class/vectors/Lesson-2/… .The horizontal and vertical motions of a projectile are independent of each other. – user81619 Aug 24 '15 at 18:10

The ball was in flight for four seconds: we can safely say that the ball reached maximum height at $t = 2$. (The gravitational pull is constant and there are no other forces acting, so the flight path is symmetrical). The ball was stationary at $t=2$ so its speed is $=0$

So now use the formula $v= u + at$, where $a$= acceleration, $t$= time, $u$= initial speed, $v$= final speed:

$v= u + at$

$0 = u - (g)(2)$

therefore $u = 2g$

And so we have the initial upwards speed component.

We could calculate the velocity at which it lands, but it's easier to argue that when it lands it will also land at the same speed due to conservation of energy. To work out the maximum height you could use the formula:

$h = ut + (1/2)(a)(t^2)$ when $t=2$, where the variables represent the same quantities.

The maximum height gained is in fact $19.6m$ when you plug in $u=2g$, $a=9.8$ .

Note that the $30m/s$ is irrelevent since it is the HORIZONTAL speed which has no effect on the vertical speed.

• Thanks, I'm new to this stuff and have lots to learn. :) – Hakkihan Tunbak Aug 25 '15 at 3:09
• This answer is not an answer to the OP's question. It's the solution to another problem, namely where the golf ball is projected from the ground and with both a vertical and horizontal speed component. But the question states clearly "Calculate the height the golf ball was projected from", so the golf ball wasn't projected from the ground. This answer also doesn't address the second question of the OP: "Calculate the magnitude of golf ball's vertical velocity component just before hitting ground". This should not have been upvoted in my opinion. – Gert Aug 25 '15 at 12:43

$\Delta y=78.4m$ is the correct answer.

The golf ball was fired horizontally from a height of $78.4m$.

With $\Delta y= \frac{1}{2}gt^2$ and $t=4s$ we get $\Delta y=78.4m$.

The horizontal velocity component of $30m/s$ was a distraction, not needed for calculating the solution because the vertical component and horizontal component of velocity are completely independent of each other.

The $30m/s$ would allow you to calculate how far the golf ball flew before hitting the floor but that wasn't part of the problem.

As regards the golf ball's vertical velocity component just before hitting the ground, it's given by $v_y=gt$, thus $v_y=39.2m/s$.