I think we will use the law of conservation of linear momentum;
M1V1 = M2V2
M1 = 4 kg (mass of the water balloon launcher)
V1=?
M2= 0.5 kg ( mass of the balloon)
V2 = 3 m/s
Therefore; 4 V1 = 0.5 × 3
4V1= 1.5
V1= 1.5/4
= 0.375 m/s
Answer:
ms⁻¹
Explanation:
= diameter of merry-go-round = 4 m
= radius of merry-go-round =
=
= 2 m
= moment of inertia = 500 kgm²
= angular velocity of merry-go-round before ryan jumps = 2.0 rad/s
= angular velocity of merry-go-round after ryan jumps = 0 rad/s
= velocity of ryan before jumping onto the merry-go-round
= mass of ryan = 70 kg
Using conservation of angular momentum



ms⁻¹
1) In a circular motion, the angular displacement

is given by

where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire:

. Using the information about the radius,

, we find the total angular displacement:

2) If we put larger tires, with radius

, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
Answer:
Vy = V sin theta = 30 * ,574 = 17.2 m/s
t1 = 17.2 / 9.8 = 1.76 sec to reach max height
Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m
H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m
Time to fall from zero speed to ground = rise time = 1.76 sec
Vx = V cos 35 = 24.6 m / sec horizontal speed
Time in air = 1.76 * 2 = 3.52 sec before returning to ground
S = 24.6 * 3.52 = 86.6 m