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

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 2

Answer:

Part a)

$t = 0.48 s$

Part b)

$x = 1.5 m$

Part c)

$v = 8.23 m/s$

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

$y = y_o + v_y t + \frac{1}{2}gt^2$

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

so we have

$t = 0.48 s$

Part b)

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

$x = v_x t$

$x = 3.19 \times 0.48$

$x = 1.5 m$

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

$v = \sqrt{v_x^2 + v_y^2}$

$v = 8.23 m/s$