Question

A juggler throws a beanbag into the air with a speed of 1.0
How long does it take for the beanbag to reach its maximum height?
Round the answer to two significant digits.
Answer using a coordinate system where upward is positive.
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Answer 1

Answer:

It takes for the beanbag 5/49 seconds to reach its maximum height

Explanation:

A juggler throws a beanbag into the air with a speed of 1.0

The initial speed is 1 m/s

The  beanbag will reach to its maximum height

At the maximum height the speed equal to zero

The final speed is zero

The acceleration of gravity = -9.8 m/s

We need to find how long it takes for the beanbag to reach its maximum

height

Lets use the rule: v = u + gt, where v is the final speed, u is the initial

velocity, t is the time and g is the acceleration of gravity

v= 0 , u = 1 , g = -9.8

0 = 1 - 9.8 t

Add two sides by 9.8 t

9.8 t = 1

Divide both sides by 9.8

t = 5/49 seconds

It takes for the beanbag 5/49 seconds to reach its maximum height

Answer 2

When a juggler throws a beanbag into the air with a speed of 1.0 m/s, it will take 0.10 seconds to reach its maximum height.

The time that takes the beanbag to reach the maximum height can be calculated with the following equation:

$v_{f} = v_{i} - gt$

Where:

$v_{f}$: is the final speed = 0 (at the maximum height)

$v_{i}$: is the initial speed = 1.0 m/s

g: is the acceleration due to gravity = 9.81 m/s²

t: is the time =?

Hence, the time is:

$t = \frac{v_{i}}{g} = \frac{1.0 m/s}{9.81 m/s^{2}} = 0.10 s$

Therefore, the beanbag will reach the maximum height in 0.10 seconds.

You can find more about maximum height here:

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I hope it helps you!