Physics questions please help!?

Question

1) A ball is thrown at an angle 45° above the horizontal with an initial velocity of 20 m/s. What is the total time of flight of the ball before it hits the ground?

2)A projectile is launched from ground level with a certain speed. For any range less than the maximum range there are two possible launch angles that give the same range. True OR False?

3) A hockey puck slides off the edge of a table with an initial velocity of 20.0 m/s. The height of the table above the ground is 2.00 m. How far from the edge of the table, measured along the floor, does the puck hit the floor?

4)A person throws a ball horizontally from the top of a building that is 40.0 m high. The initial velocity of the ball is 100 m/s. What is the horizontal distance that the ball travels before hitting the ground?

Answer 1

#1)

Voy = Vo*sin(45) = 14.142m/s

Vf = Vo + at

t = (Vf - Vo) / a

t = (0 - 14.142) / (-9.8) = 1.443s

**this t-value is known as the "half-time flight." 1.443s until there is no more force in the y-axis, which means the ball has reached it maximum height. because this t-value only deals with the first half of it's flight, you must multiply it by 2 to get the "full-flight time" **

t = 1.443s * 2 = 2.89s <--- answer

#2) False. anything above 45 degrees travels higher and for a shorter distance, anything below 45 degrees travels lower and for a longer distance. 45 degrees. No two different angles can give the same range.

#3)

we can use the height and acceleration to solve for the time, then use the time to solve for the distance from the edge of the table.

(y-direction)

x = Vo*t + (1/2)a * t²

2m = 0t + (1/2)(9.8) * t²

2m = 4.9t²

t = √(2/4.9) = 0.639s

now we can use the time to solve for distance from the table:

**remember, there is no acceleration in the x-direction**

(x-direction)

x = Vo*t  + (1/2)at²

x = (20m/s)(0.639) + (1/2)(0)(0.639s)²

x = (20m/s)(0.639) + 0

x = 12.78m  <---answer

#4)

we can use the exact same method to solve this problem.

we will find the time, then use the time to find the distance.

(in the y-direction)

x = Vo * t + (1/2)at²

40m = 0 + (1/2)(9.8) t²

40m = 4.9t²

t = √(40 / 4.9) = 3.714s

(in the x-direction)

x = Vo * t + (1/2)at²

x = (100m/s)(3.714s) + (1/2)(0)(4.714s)²

x = (100m/s)(3.714s) + 0

x = 371.4m <---answer

i hope this helped you!