When a stone is thrown directly upwards with initial velocity of 30m s what will be the maximum height it will reach and when will it be?

Time taken by stone to rise to its maximum height is T=ug=3010=3 sec.

What is the time at the maximum height if the initial velocity is 30 m s?

in our case: 0=30−9.8t the acceleration is negative because it is directed downwards (see the diagram). So you have that the time to reach the maximum height will then be: t=309.8=3s .

How to calculate the maximum height of an object thrown at an angle?

The maximum height reached can be calculated by multiplying the time for the upward trip by the average vertical velocity. Since the object’s velocity at the top is 0 m/s, the average upward velocity during the trip up is one-half the initial velocity.

At what angle should an object be thrown to reach the maximum height given that the air resistance is neglected?

45 degrees
A projectile, in other words, travels the farthest when it is launched at an angle of 45 degrees.

When a stone is thrown directly upwards with initial velocity of 30m S?

The initial velocity is 30 m/s, and the velocity drops by 9.8 m/s each second due to the pull of gravity. Maximum height is at the moment the stone has zero velocity. t = v/a = 30/9.8 = 3.06 seconds.

When a ball is thrown vertically upwards at the maximum height?

Hint: When a ball is thrown upwards with a certain velocity, the ball moves against the acceleration due to gravity and attains a maximum height. At the point of maximum height, the velocity of the ball is zero.

When a ball reaches maximum height What is the velocity?

0 m/s
When a projectile reaches maximum height, the vertical component of its velocity is momentarily zero (vy = 0 m/s).

What is the formula for maximum height?

h = v 0 y 2 2 g . h = v 0 y 2 2 g . This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical component of the initial velocity.

What is the maximum height equation?

The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 ⁡ θ i 2 g . The time of flight of an object, given the initial launch angle and initial velocity is found with: T=2visinθg T = 2 v i sin ⁡ .

What is the angle of projection for maximum range?

The textbooks say that the maximum range for projectile motion (with no air resistance) is 45 degrees.

At what angle range is maximum?

The range of a projectile for a given initial velocity is maximum when the angle of projection is 45∘.

How do you find the magnitude of projectile motion?

along the horizontal and vertical axes. Its magnitude is s and it makes an angle θ with the horizontal. To describe projectile motion completely, we must include velocity and acceleration, as well as displacement. We must find their components along the x- and y -axes.

What is the maximum height a projectile can reach?

Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance).

What happens to the vertical velocity as an object rises?

(c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity.

What are the applications of projectile motion in physics?

The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as they enter Earth’s atmosphere, fireworks, and the motion of any ball in sports. Such objects are called projectiles and their path is called a trajectory.