To solve this problem you use the derived formula ∆dh=(V1h2(sin2Ѳ))/g 250m=V1h2(sin2(17))/9.8m/s2 V1h=((sin(17))/9.8m/s2)1/2 V1h= 64.7292201m/s V1h= 65m/s
To solve this problem you use the derived formula ∆dh=(V1h2(sin2Ѳ))/g 250m=V1h2(sin2(17))/9.8m/s2 V1h=((sin(17))/9.8m/s2)1/2 V1h= 64.7292201m/s V1h= 65m/s
its 45 degree
A degree is an angular measure and cannot be measured in millimetres. A 1 degree rise can be interpreted as a ratio of a rise (in millimetres) per a distance of horizontal displacement.
If you ignore the effect of the air grabbing at it and only figure in gravity, then the horizontal component of velocity is constant, from the time the stone leaves your hand until the time it hits the ground. Makes no difference whether you toss it up, down, horizontal, or on a slant. Also makes no difference whether it's a cannonball, a stone, or a bullet.
Both the arrows will cover equal horizontal distance as their angles are at an difference from 45 degree. Horizontal range is maximum at 45 degrees and decreases equally on sum or difference of an angle from 45. But vertical distance increases on addition of an angle from 45 and decreases on subtraction of angle from it. For more details, contact at saqibahmad81@yahoo.com
The answer is 0.33
Horizontal position is strate 90 degree
It will increase until you hit 45 degrees and the it will begin to drop.
4
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?
yes it does because you can get the maximum distance from a 45 degree angle - at an elevation of 0, but as your beginning height (above the surface) increases the angle for maximum range changes.
90 degree