It takes 5 seconds to reach the top of its path.
8.15second
The time taken by the ball to reach the maximum height is 1 second. The maximum height reached by the ball is 36 meters.
A ball is thrown vertically upward with an initial speed of 20m/s. Two second later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24m/s. At what height above the release point will the ball and stone pass each other?
0.82 metres.
That depends on a number of different variables and therefore it cannot be concluded here. It depends on the mass of the object being swung as well as the initial conditions of this object such as the height it is released or the initial velocity by which it was flung.
If the object has less speed, then it will fall back to Earth.
The time taken by the ball to reach the maximum height is 1 second. The maximum height reached by the ball is 36 meters.
If the initial velocity is 50 meters per second and the launch angle is 15 degrees what is the maximum height? Explain.
The rocket would attain a maximum height of 158.65 feet (63.65 feet from the top of the structure).
1.6 ft
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.
23 sec
The maximum height the water will reach is 157 m.
A ball is thrown vertically upward with an initial speed of 20m/s. Two second later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24m/s. At what height above the release point will the ball and stone pass each other?
the second space probe launched was called the kooterliker01
0.82 metres.
0.82 metres.
Assuming the simple model where the object is projected with an initial velocity of u metres/second at an angle of x to the horizontal, and that the only force acting on it after that is gravitational acceleration, g = 9.81 metres/second^2, then h = [u*sin(x)]^2/(2*g) metres.If the launch is vertical then x = pi/2 radians and h = u^2/(2*g) metres.