(Once again, as we always do with problems like this one, we have to ignore
the effects of air resistance. Sorry. It's just too complicated if we don't.)
The acceleration of gravity is 9.8 meters per second2. That means that any
object that's not propelled or constrained gains 9.8 m/s of downward speed
every second.
The arrow starts with 196 m/s of upward speed. It loses 9.8 m/s of that speed
every second. It'll run out of upward speed and start moving downward after
196/9.8 = 20 seconds .
The time taken by the ball to reach the maximum height is 1 second. The maximum height reached by the ball is 36 meters.
The horizontal component of a projectile's velocity doesn't change, until the projectile hits somethingor falls to the ground.The vertical component of a projectile's velocity becomes [9.8 meters per second downward] greatereach second. At the maximum height of its trajectory, the projectile's velocity is zero. That's the pointwhere the velocity transitions from upward to downward.
It depends. If the projectile goes straight up and straight down, its velocity will be zero at the top. If the projectile is a baseball about halfway between the pitcher and the bat, its velocity might be 150 km/h.
In a vacuum it would fall back to the same height at the same speed, 150m/s. It would then gain another minuscule fraction of speed as it fell from bat-height down to the ground. In the real world, a falling baseball will reach a maximum speed of around 42 meters per second because the air resistance slows it down. This is called Terminal Velocity.
The maximum speed at which grenade shrapnel can travel upon explosion is typically around 1,600 meters per second.
Initial upward speed = 7.61 m/sFinal upward speed (at the point of maximum height) = 0Time to reach maximum height = (7.61) / (9.8) = 0.77653 secondAverage speed during that time = 1/2 ( 7.61 + 0) = 3.805 m/sHeight = 3.805 x 0.77653 = 2.9547 meters (rounded) = about 9.7 feetDoesn't seem like much of a height for a strong toss; but the math looks OK.
The speed of an object falling from a great height is measured in meters per second per second until it reaches terminal velocity (maximum downward speed).
If the initial velocity is 50 meters per second and the launch angle is 15 degrees what is the maximum height? Explain.
it is 10 meters per second straight down
It depends on the height of the building and also on the direction the object is thrown in (up, down etc.).
The time taken by the ball to reach the maximum height is 1 second. The maximum height reached by the ball is 36 meters.
18
the second tallest is mount god-win austen and its height is 28251 feet (8611 meters)
The maximum height of the ball above the ground can be calculated using the vertical component of the initial velocity. Assuming no air resistance, the formula to determine maximum height is h = (v^2 sin^2(theta)) / (2g), where v is the initial velocity (16 m/s), theta is the angle (40 degrees), and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, you can find that the maximum height of the ball is approximately 14.1 meters.
The second-highest mountain in the world is K2 with a height of 8,611 meters.
The second-highest mountain in the world is K2 with a height of 8,611 meters.
It is always -9.8 meters per second squared, regardless of height