# Is there a difference between initial speed and initial velocity?

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1. Nov 19, 2014

### ncredibler

If a projectile is thrown at an angle of 60o and it's initial speed is 30m/s, find the maximum height, time used for the ball to reach the ground, the horizontal range the ball reached from starting point, and the final velocity.

So can i use 30m/s as the initial velocity?

2. Nov 19, 2014

### Drakkith

Staff Emeritus
Technically 30 m/s is the speed. Velocity is specifying the speed and the direction both, which is 30 m/s along the 60 degree angle. Speed is the magnitude of the velocity and is a scalar quantity (meaning it only takes one number to describe it, unlike a vector which takes at least two, one for the magnitude and one for the direction).

3. Nov 19, 2014

### Staff: Mentor

You can use 30 m/s as the magnitude of the initial velocity. Velocity is a vector quantity, so to specify the intial velocity you must provide not only its magnitude (i.e., the speed) but also a direction.

So the initial velocity you are concerned with here is 30m/s at +60° to the horizontal.

4. Nov 19, 2014

### ncredibler

okay... hmm..
okay... so I'll have to calculate for the initial velocity right?

5. Nov 19, 2014

### NTW

That 'problem' exist only in English. In other languages, like French, German or Spanish, only 'vitesse', 'Geschwindigkeit' and 'velocidad' are used, both as vectors or scalars. When one also sees 'rapidité, 'Schnelligkeit' or 'rapidez', it's usually in books translated from English.

6. Nov 19, 2014

### rtsswmdktbmhw

You don't need to calculate initial velocity - it is 30 m/s at 60 degrees. But to solve your problems, you will need to calculate the vertical and horizontal components of the initial velocity.

7. Nov 19, 2014

### ncredibler

Alright :) thanks

8. Nov 19, 2014

### ncredibler

So far this is how I've attempted to solve the problems
Initial Velocity
Vox=Vicosθ=(30m/s)(cos60o)=15m/s
Voy=Visinθ=(30m/s)(sin60o)=26m/s

Time used for ball to reach the ground
ΔVy=Vy-Voy=Voyt-1/2(g)(t2)
ΔVy=0=(30m/s)(sin60)=1/2(9.8m/s2)
=5.3s

Range
Range=Vxt=(30m/s)(cos60o)(5.3s)
=79.5m

Maximum Height
Vymax=(Vy+Voy)t/2
Vymax=(0+30sin60o)(2.65)/2
=34.4m

Can anyone help me to see if my calculations are correct? and how can I get the final velocity? :/

9. Nov 19, 2014

### lep11

You should use letters x and y for displacement and v or |v| for speed. Your calculations are okay except the highlighted part. Maybe there are steps missing. The ball will reach its maximum height when t=5.3s/2 because the projectile is parabolic and symmetric so substitute t=5.3s/2 into y=v0yt-(1/2)gt2 and you will get the maximum displacement in y-direction.

In x-direction vector v0x remains the same (magnitude and direction won't change) if we assume there is no drag. You know how to calculate the magnitude of v0x

In y-direction

At the instant the ball reaches its highest point the speed in y-direction will be zero. Then the direction of vy will change towards the ground and the magnitude of that velocity vector will increase until the ball hits the ground. That acceleration is obviously caused by the gravitational force acting on the ball. Assuming g is constant, the magnitude of vector vy at time t will be vy=v0y-gt

You know the head-to-tail method for vector addition so the final velocity will be the resultant of these two vectors. Vectors are usually written in bold.

Last edited: Nov 19, 2014
10. Nov 20, 2014

### ncredibler

Okay so max height:
y=v0yt-(1/2)gt2=(26m/s)(5.3s/2)-(1/2)(9.8m/s)(5.3/2)2
=34.4 m

For final velocity:
Vfx=15m/s because ax=0
Vfy=v0y-gt=26-(9.8)(5.3)=-25.94
Vf=√(152+(-25.94)2)=29.93m/s
Vf=29.93m/s 60o at the horizontal

Have I done it right?

Last edited: Nov 20, 2014
11. Nov 20, 2014

### lep11

Looks fine :)

Last edited: Nov 20, 2014
12. Nov 20, 2014

### ncredibler

yey! thanks :D