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.ncredibler said: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?
Drakkith said: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).
Drakkith said: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).
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.ncredibler said:okay... so I'll have to calculate for the initial velocity right?
rtsswmdktbmhw said: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.
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.ncredibler said: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? :/
Okay so max height:lep11 said: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.
Looks fine :)ncredibler said: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/s
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?
yey! thanks :Dlep11 said:Looks fine :)
The main difference between initial speed and initial velocity is that speed is a scalar quantity, meaning it only has a magnitude, while velocity is a vector quantity, meaning it has both magnitude and direction. In other words, initial velocity takes into account the direction of motion, while initial speed does not.
No, initial speed and initial velocity are only equal if the motion is in a straight line. If the motion involves a change in direction, the initial speed and initial velocity will be different.
Yes, initial speed and initial velocity can have different values if the motion involves a change in direction. For example, if an object is thrown upwards, its initial speed will be the same as its final speed (when it reaches the highest point), but its initial velocity will be different as it changes direction from upwards to downwards.
Initial speed and initial velocity are typically measured in units of distance per time, such as meters per second (m/s) or kilometers per hour (km/h). These units represent the magnitude of the speed or velocity. To measure the direction of initial velocity, a reference point or axis is needed.
Understanding the difference between initial speed and initial velocity is important in accurately describing and predicting the motion of objects. For example, in the field of physics, velocity is used to calculate important quantities such as acceleration and displacement, while speed cannot be used in these calculations. Additionally, in real-world situations, initial velocity plays a crucial role in determining the trajectory and final position of a moving object.