Question

From the roof of a tall building, a student throws a ball of mass m = 0.2 kg with initial speed vo = 30 m/s at an angle θ = 20o below the horizontal as shown in the figure. The ball is thrown from a height H above the ground. The student measures the ball to be in the air for 5 seconds before it hits the ground. Ignore air resistance.

What is D, the horizontal distance from the building to where the ball hits the ground?

Answer 1

Break your initial speed into it's "x" and "y" components by using the angle. Then use the time that the ball is in air and plug it into your kinematic equation for displacement with time equation:

horizontal distance = (v)(t)

Answer 2

i end up with 263.45 which is wrong. (30*cos(20))*5 + 4.9(25)

Answer 3

where are you getting 4.9(25) from?

Answer 4

nevermind, i read the equation wrong. how about this problem?

A box of mass m = 3 kg is being pulled by a horizontal string across the top of a second box having a mass M = 5 kg. The kinetic coefficient of friction between the upper box and the lower box is μ = 0.25. There is no friction between the lower box and the floor. The upper box moves with a constant velocity.

Answer 5

What's the question?

Answer 6

What is aM,x, the x-component of the acceleration of the bottom block?

Answer 7

Draw a Free Body diagram of each box then apply Newton's Third law at the interaction between each box from there you can find the acceleration of box "M" by diving the Force that is applied by the kinetic friction between the boxes.

Use your kinetic friction force formula:
\[F _{\mu _{k}}= \mu _{k}N\] where "N" is your normal force. Find that also from the mass of the box "m"