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.
A rock is thrown upward at an angle of 50 degrees with respect to the horizontal component of velocity remains unchanged. The vertical component decreases.
When a rock is thrown upwards at an angle, it starts to slow down on its horizontal velocity because of gravitational pull. The stone can then start to accelerate again as it nears the ground.
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.
You can solve this using kinematics: Since you have height: 20m and acceleration being 10 for this question you can easily find the final velocity The initial velocity MUST be 0 since you are dropping it. with that being said Use: V^2= Vi^2 + 2ad
You can't find the resultant of two vectors without magnitudes as well as directions.
Assuming position is measured between the cars and not their starting points?With constant speed the Distance over time graph would have a 1:1 slope or 45 degree angle.While the Velocity (which is constant) over time would be be a horizontal flat line.
For 1 degree Celsius rise in temperature in air the speed of sound goes up by 0.60 m = 60 cm.
You have a right triangle and can use trig. Degree mode. tan(theta) = adjacent/opposite( y component ) tan( 60 degrees) = (5 m/s)/(y comp.) y component = 5 m/s)/(tan 60 degrees) = 2.887 m/s ( you can call it 3 m/s ) -----------------------------------------------
-- The component that's inclined 30 degrees above the horizontal is[ 20 sqrt(3) ] = about 34.641 newtons. (rounded)-- The other component is inclined 60 degrees below the horizontal,and its magnitude is 20 newtons.
Horizontal position is strate 90 degree
RESPECT RESPECT
4
Suppose a projectile is fired from a gun, we know that "g" remains constant and as we use horizontal component of velocity in range sov0 also remains constant. Only sin2θ responsible for change in range. The range will be maximum if sin2θ has its maximum value that is 1.for maximum range:sin2θ = 12θ = sin-1 (1)θ = 90/2θ = 45 (degree)therefor if projectile is projected with the angle of 45(degree) its range will be maximum.
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?
90 degree
It doesn't matter how long she maintains her acceleration.As long as she continues to accelerate at 1.2 m/s2 parallel to the slope,the vertical component of her acceleration is1.2 sin(11°) = 0.229 m/s2 (rounded)and the horizontal component is1.2 cos(11°) = 1.178 m/s2(rounded)
A velocity time graph is still a velocity time graph - no matter the degree of detail that you look at it.
Ask Sir JB.
An obtuse angle is an angle between 90 degrees and 180 degrees. A 180 degree angle is a straight horizontal line. So you can imagine that a 160 degree angle is almost but not quite horizontal.