Question

A friend tosses a baseball out of his second floor window with initial velocity of 4.3m/s(42degrees below the horizontal). The ball starts from a height of 3.9m and you catch the ball 1.4m above the ground.

a) Calc the time the ball is in the air (ans. 0.48s)
b)Determine your horisontal distance from window (ans. 1.5 m)
c)Calc the speed of ball as you catch it (ans: 8.2m/s)

I dont get what 42 m below the horizontal is, can someone give me direction on how to do this?

Answer 1

When the initial velocity of the baseball is stated as 4.3 m/s at 42 degrees below the horizontal, it means the ball is being thrown downward at an angle. To solve the problem, you need to break down the velocity into horizontal and vertical components.

Step-by-step explanation:

When it is stated that the initial velocity of the baseball is 4.3 m/s at 42 degrees below the horizontal, it means that the ball is being thrown with a velocity of 4.3 m/s at an angle of 42 degrees below the horizontal line. The horizontal line here represents a flat surface.

To solve the problem, you can break down the initial velocity into its horizontal and vertical components. The horizontal component represents the ball's motion parallel to the ground, while the vertical component represents the ball's motion in the vertical direction.

In this case, the ball is thrown downward, so it has a negative vertical component of velocity. The negative sign indicates that the ball is moving downward instead of upward.

Answer 2

Part a)

`t = 0.48 s`

Part b)

`x = 1.5 m`

Part c)

`v = 8.23 m/s`

Step-by-step explanation:

As we know that the velocity of ball is

`v = 4.3 m/s`

now the two components of velocity is given as

`v_x = 4.3 cos42 = 3.19 m/s`

`v_y = 4.3 sin42 = 2.88 m/s`

Part a)

now in Y direction we will have

`y = y_o + v_y t + (1)/(2)gt^2`

`1.4 = 3.9 - 2.88 t - 4.9 t^2`

so we have

`t = 0.48 s`

Part b)

Now the distance covered by the ball in horizontal direction is given as

`x = v_x t`

`x = 3.19 * 0.48`

`x = 1.5 m`

Part c)

speed in x direction will always remain the same

so we have

`v_x = 3.19 m/s`

speed in y direction is given as

`v_y = v_i + at`

`v_y = 2.88 + (9.8)(0.48)`

`v_y = 7.58 m/s`

So final speed will be

`v = √(v_x^2 + v_y^2)`

`v = 8.23 m/s`