Question

A ball player catches a ball 3.2 s after throwing it vertically upward. what height did it reach?

Answer 1

The ball did not reach any height as it fell straight back down to the player.

Step-by-step explanation:

To find the height the ball reached, we can use the equation for the motion of an object thrown vertically upwards. The equation is given by:

h = (v02 - v2)/2g

Where:

• h is the height
• v0 is the initial velocity
• v is the final velocity (0 m/s at the highest point)
• g is the acceleration due to gravity (9.8 m/s2)

Since the ball was caught, we know that its final velocity is 0 m/s. We also know that it took 3.2 seconds to reach that point. Substituting these values into the equation, we get:

h = (v02 - 0)/2(9.8)

Simplifying the equation, we have:

h = v02/19.6

To find the value of v0, we can rearrange the equation:

v0 = sqrt(h * 19.6)

Substituting the given value of h, we find:

v0 = sqrt(0 * 19.6) = 0 m/s

Therefore, the ball did not reach any height as it fell straight back down to the player.

Answer 2

At the top of the height, the velocity is zero and acceleration is negative of acceleration due to gravity ( i.e
`-9.8 m/s^2`).

The time of the ball in air is 3.2 s, so ascending time is
`3.2/2=1.6 s`.

Therefore from kinematic equation,

`v = u + gt`

Substituting the values we get,

`0= u - 9.8 (1.6)\\\\u=15.7 m/s`, Here v = 0 at top.

Now from equation,

`h=ut+(1)/(2) gt^2`, here h is the height .

So,

`h=(15.7 m/s) (1.6s)-(1)/(2) 9.8m/s^2(1.6)^2\\\\ h=25.12m-12.54m\\\\h=12.48 m`.

Thus, the ball reached at its maximum height of 12.48 m.