x is horizontal, y is vertical
v=11m/s
z=angle=33°
vy=v*sin(33°)=5.99m/s
vx=v*cos(33°)=9.22m/s
we have to calculate how long the object will be in the air, the best way to calculate this is to caltulate how long does it take for the object to reach the maximum. we take it as we were to throw the object vertically up at vy speed.
v=vy + a*t v=0, because it stops at the max, so:
vy=-a*t (but im gonna leave out the minus, because we wanna know just the number)
we wanna know how much does it take for the object to reach the highest point
t1=vy/a = 0.61s (a is g=9.81m/s^2)
so tha total time of the object in the air is t=2*t1=1.22s
the vx speed does not vary with time, because there is no acceleration in the horizontal axis, so how far does the object get? v=s/t -> s=vx*t =11.25m and that is the result
LOOPDOP says: More accurately: Distance = v2 / g * sin (2 * angle)
= 112 / 9.81 * sin 66 degr. = 11.268m (rounded)
That is the distance you asked which is also the max. range.
To increase range you have to throw the stone with higher velocity than 11m/sec at 33o launch angle.
How far? In absence of air resistance, distance will be 24.522 meters (rounded). The max. range that will be achieved will depend on how much faster you can throw that stone but for 16m/sec initial velocity it will only reach 24.522m at 35 degr. launch angle.
A stone is thrown with an angle of 530 to the horizontal with an initial velocity of 20 m/s, assume g=10 m/s2. Calculate: a) The time it will stay in the air? b) How far will the stone travel before it hits the ground (the range)? c) What will be the maximum height the stone will reach?
Making the improbable assumption that the jumper experiences no air resistance, he will jump 3.97 metres, and reach a height of 0.72 metres.
It is simply a plane surface making an angle with the horizontal (ground).
Maximum of one.
A projectile has maximum horizontal range when it is launched at an angle of 45 degrees to the horizontal. This angle allows for the ideal balance between the horizontal and vertical components of the projectile's velocity, ensuring that it travels the farthest distance before hitting the ground.
50m
No, a ball thrown at an angle of around 45 degrees will typically go the furthest. This launch angle combines vertical and horizontal components of velocity to achieve the maximum range. A ball thrown almost horizontally will have a lower trajectory and be affected more by air resistance, causing it to have a shorter range.
The maximum range of a projectile is the distance it travels horizontally before hitting the ground. It is influenced by factors such as initial velocity, launch angle, and air resistance. In a vacuum, the maximum range is achieved at a launch angle of 45 degrees.
How far? In absence of air resistance, distance will be 24.522 meters (rounded). The max. range that will be achieved will depend on how much faster you can throw that stone but for 16m/sec initial velocity it will only reach 24.522m at 35 degr. launch angle.
Horizontal markings are markings (such as painted or etched lines) that are horizontal (or parallel to the ground).
Air moving parallel to the ground is called horizontal wind or surface wind.
Horizontal markings are markings (such as painted or etched lines) that are horizontal (or parallel to the ground).
the word horizontal means: flat or level: a horizontal position. "The line is horizontal to the ground" would work.....
If the non-horizontal projectile is launched abovehorizontal, thenit's the second one to hit the ground, after the horizontal one.If the non-horizontal one is launched below horizontal, then it'sthe first to hit the ground, before the horizontal one.
A stone is thrown with an angle of 530 to the horizontal with an initial velocity of 20 m/s, assume g=10 m/s2. Calculate: a) The time it will stay in the air? b) How far will the stone travel before it hits the ground (the range)? c) What will be the maximum height the stone will reach?
Sheep and goats have horizontal, oval pupils.