Place where warm air rises and pushes down on the earths surface with less pressure

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

cluponka [151]

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,

$h=ut+\frac{1}{2} gt^2$ , here h is the height .

So,

$h=(15.7 m/s) (1.6s)-\frac{1}{2} 9.8m/s^2(1.6)^2\\\\ h=25.12m-12.54m\\\\h=12.48 m$ .

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

8 0

3 years ago

Assume it takes 8.00 min to fill a 50.0-gal gasoline tank. (1 U.S. gal = 231 in.3) (a) Calculate the rate at which the tank is f

vagabundo [1.1K]

The volumetric rate or flow rate of a fluid is defined as the amount of the volume of a fluid circulating on a surface per unit of time. In this case we have units given initially: Gallons and minutes. For the first part we will convert the minutes to seconds, and we will obtain the flow rate under that measure. For the second case we will convert the gallons to cubic meters and obtain the desired value. Recall the following conversion rates,

$1 min = 60s$