A ball rolls along a desktop with an uniform velocity of 3 m/s. What is the displacement after 10 seconds have passed?

A ball is thrown upward with an initial velocity of 18 m/s? What is the maximum height the ball reaches if the ball is assumed to leave from the ground?


A ball is dropped from a 35 m tall building. The ball takes how long to impact the ground?

1. D = Vt = 3m/s + 10s = 30m.

2. h = (Vf^2 - Vo^2) / 2g,
h = (0 - (18)^2) / -19.6 = 16.53m.

3. d = Vo*t + 0.5g*t^2 = 35m.
0 + 4.9t^2 = 35,
t^2 = 7.14,
t = 2.67s.

To find the displacement of the ball after 10 seconds, we can use the formula: displacement = velocity × time. In this case, the velocity of the ball is 3 m/s and the time is 10 seconds. Therefore, the displacement would be 3 m/s × 10 s = 30 m. So, the ball would have a displacement of 30 meters after 10 seconds.

To find the maximum height reached by the ball when thrown upward, we can use the equation for vertical motion under constant acceleration. The initial velocity is 18 m/s and the acceleration due to gravity is -9.8 m/s² (negative because it acts against the upward motion). The maximum height can be found using the formula: maximum height = (initial velocity²) / (2 × acceleration). Plugging in the values, we get: maximum height = (18 m/s)² / (2 × -9.8 m/s²) = 32.65 m. Therefore, the maximum height reached by the ball is approximately 32.65 meters.

To find the time it takes for the ball to impact the ground when dropped from a 35 m tall building, we can use the equation for vertical motion under constant acceleration. The initial velocity is 0 m/s since the ball is dropped and the acceleration due to gravity is -9.8 m/s² (negative because it acts in the opposite direction of upward motion). We can use the formula: time = √(2 × height / acceleration). Plugging in the values, we get: time = √(2 × 35 m / -9.8 m/s²) = √(70 m / -9.8 m/s²) = √(-7.14 s²) = 2.67 s (approx.). Therefore, it takes approximately 2.67 seconds for the ball to impact the ground.