A kangaroo leaps upward at time t = 0.  At time t = 1.0 s it is at a height of 2.0 m above the ground. At what time does it reach its maximum height?Express your answer in seconds. (Assume proximity to the Earth's surface, and neglect friction).

To solve this problem, we need to understand the physics of a projectile motion. When an object is thrown upwards, it will rise until the force of gravity pulls it back down. The time it takes to reach the maximum height is when the velocity of the object becomes zero.

Given that the kangaroo jumps up at t=0 and at t=1.0s it is at a height of 2.0m, we can use the equation of motion to find the initial velocity (u) of the kangaroo.

The equation of motion is: h = ut + 0.5gt^2, where h is the height, u is the initial velocity, g is the acceleration due to gravity (which is approximately 9.8 m/s^2), and t is the time.

We know that h=2.0m, t=1.0s, and g=9.8 m/s^2. Substituting these values into the equation, we get:

2.0 = u1.0 + 0.59.8*(1.0)^2

Solving for u, we get u = 2.0 - 0.59.81.0 = 2.0 - 4.9 = -2.9 m/s. The negative sign indicates that the velocity is directed upwards.

Now, we can find the time it takes for the kangaroo to reach its maximum height by setting the velocity to zero and solving for t. The equation for velocity is v = u - gt.

Setting v=0, we get 0 = -2.9 - 9.8t.

Solving for t, we get t = -2.9 / -9.8 = 0.296 s.

So, the kangaroo reaches its maximum height at t = 0.296 s after it starts jumping.

To solve this problem, we need to understand the physics of a projectile motion. When the kangaroo jumps, it initially has an upward velocity which gradually decreases due to the force of gravity. When the kangaroo reaches its maximum height, its velocity becomes zero for an instant before it starts falling back down.

We can use the following kinematic equation to solve this problem:

v = u - gt

where: v is the final velocity (0 m/s at the maximum height), u is the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time.

We don't know the initial velocity, but we can find it using the given information. At t = 1.0 s, the kangaroo is at a height of 2.0 m. We can use the equation of motion:

h = ut - 0.5gt²

Substituting the given values:

2.0 m = u(1.0 s) - 0.5(9.8 m/s²)(1.0 s)² 2.0 m = u - 4.9 m u = 6.9 m/s

Now we can find the time it takes for the kangaroo to reach its maximum height by setting v = 0 in the first equation and solving for t:

0 = 6.9 m/s - 9.8 m/s² * t 9.8 m/s² * t = 6.9 m/s t = 6.9 m/s / 9.8 m/s² t = 0.704 s

So, the kangaroo reaches its maximum height approximately 0.704 seconds after it starts jumping.