Pllllllz help me ineed to hand it today

[night rose]

please help how can i solve these qustion i should send it to the prof today

1)a ball was thrown horizontally with an initial velocity of 20 m/s from the edge of a building of a certain high. the ball lands at a horizontal distance of 82.0m from the base of the building. What is the height of the building?


2)a bullet is fired from ground level with a speed of 150 m\s at an angle 30 degrees above the horizontal at a location where g=10.0m/ss. what is the vertical component of its velocity when it is at the highest point of it trajectory

3)a person throws a ball horizontally from the top of a building that is 40m high. the initial velocity of the ball is 100 m/s. what is the horizontal distance that the ball travels before hitting the ground.?
4)in a relay race, a runner A is carrying the baton and has a speed of 3.40m/s. when he is 25.0 m behind the starting line, runner B starts from rest to accelerate at 0.140 m/ss. how long afterwards will A catch up with B to pass the baton to B?

5)a boy jumps at a speed of 20 m/s at an angle of 25.0 degrees above horizontal. what is the horizontal component of the boys velocity.?
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[jbsteele]

Answers:1) The height of the building is 34.7 m.2) The vertical component of its velocity when it is at the highest point of its trajectory is 75 m/s.3) The horizontal distance that the ball travels before hitting the ground is 80 m.4) It will take A 15.9 seconds to catch up with B and pass the baton.5) The horizontal component of the boy's velocity is 17.6 m/s.

 

[baba89]

1) To find the height of the building, we can use the equation d = v0^2 / 2g, where d is the horizontal distance, v0 is the initial velocity, and g is the acceleration due to gravity. Rearranging the equation to solve for the height, we get h = d * 2g / v0^2. Plugging in the given values, we get h = (82.0m) * 2(9.8m/s^2) / (20m/s)^2 = 32.2m. Therefore, the height of the building is 32.2 meters.

2) To find the vertical component of the velocity at the highest point of the trajectory, we can use the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. At the highest point, the vertical velocity is 0 since the bullet is momentarily at rest before falling back down. Therefore, we can use the equation vf^2 = vi^2 + 2ad, where d is the vertical distance. Rearranging the equation, we get d = (vf^2 - vi^2) / 2a. Plugging in the given values, we get d = (0 - (150m/s)^2) / (2 * -10m/s^2) = 1125m. Therefore, the vertical component of the velocity at the highest point is 1125m/s.

3) To find the horizontal distance traveled by the ball before hitting the ground, we can use the equation d = v0t, where d is the distance, v0 is the initial velocity, and t is the time. We can also use the equation d = (1/2)at^2, where a is the acceleration due to gravity. Since there is no acceleration in the horizontal direction, we can set these two equations equal to each other and solve for t. We get t = d / v0. Plugging in the given values, we get t = (100m/s) / (9.8m/s^2) = 10.2s. Therefore, the ball travels for 10.2 seconds before hitting the ground, and the horizontal distance traveled is d = (100m/s) * (10.2s) = 1020m.

4) To