Question

Question #d8d98

Answer 1

a): `7.25` `"m"`

b): `7"m"/"s"` downward

c): `-9.8"m"/("s"^2)`

Explanation:

I'm going to assume it was thrown straight upward, since there's only one velocity equation given.

a)

To find the maximum height reached by the tennis ball, we need to find the time `t` when the velocity is `0`:

`0 = 5-10t`

`10t = 5`

`t = 0.5` `"s"`

To find its position `y` at this time, we need to integrate the velocity equation. The position equation from velocity is given by

`y = y_0 + int_0^t v_ydt = 6.0"m" + int_0^t [5"m"/"s" - (10"m"/("s"^2))t]dt`

`= 6.0"m" + (5"m"/"s")t - (5"m"/("s"^2))t^2`

Plugging in the time `0.5"s"` into this equation, its maximum height reached is

`y = 6.0"m" + (5"m"/"s")(0.50"s") - (5"m"/("s"^2))(0.50"s")^2 = color(red)(7.25)` `color(red)("m")`

b)

Now, we're asked to find and describe the object's velocity at the time `t = 1.2` `"s"`. By this, it's asking for its magnitude and direction (which is just upward or downward).

We found in part a) that its velocity is `0` at `t = 0.5` `"s"`. At all times after this, its velocity is negative and is thus moving in the downward direction.

To find the velocity, we simply plug `t = 1.2` `"s"` into the velocity equation:

`v_y = 5"m"/"s"-(10"m"/("s"^2))(1.2"s") = color(blue)(-7"m"/"s"`

The negative sign indicates that the object is traveling downward at this time. Its velocity is thus `7"m"/"s" "downward"`.

c)

For any object that is launched (near Earth's surface), it's acceleration is always `9.8"m"/("s"^2)` toward Earth's surface (most often the "downward" direction), if air resistance is neglected. The acceleration of the object at any time is thus `color(green)(-9.8"m"/("s"^2)`.

Answer 2

(a) `7.25m`
(b) `-7ms^-1`
(c) `-10ms^-2`

Explanation:

Let us assume that the tennis ball is thrown upwards (`+ve` direction) from a tree house.

Kinematic equation for the velocity is given as

`v= 5-10t` .....(1)

Compare it with the general kinematic expression

`v= u+at`.....(2)
where `u` is the initial velocity, `v` is velocity after time `t` and `a` is acceleration.

At `t=0` we get `u=5ms^-1`

(c) Also we get `a=-10ms^-2`

(a) Greatest height is reached when `"velocity "v =0`. Inserting this condition in equation kinematic equation

`v^2-u^2=2ah` ......(3)

we get
`0^2-5^2=2(-10)h`
`=>h=25/20=1.25m`
Recall that the tree house is located `6m` off the ground. As such maximum height reached is `6+1.25=7.25m`

(b) Velocity at `t=1.2s` can be found from (1)

`v_1.2= 5-10xx1.2`
`=>v_1.2= -7ms^-1`
`-ve` sign shows that the tennis ball is moving with this velocity towards ground.

Lets verify height at `t=1.2s` from equation (3)
`(-7)^2-5^2=2(-10)h_1.2`
`49-25=-20h_1.2`
`49-25=>h_1.2=-24/20=-1.2m`
The ball is `1.2m` from the tree house towards the ground. (`-ve` sign indicates on the opposite side of its initial motion)