A snow ball is thrown upward at a speed of 27 m/s. How high is the snow ball 3.5 seconds later? Also, how fast is it moving at that time?

The snow ball will achieve the height of 34.75 m and at that height, it will move with a speed of 61.3 m/s.

Given data:

The initial speed of the snow ball is, u = 27 m/s.

The time taken by the snow ball to achieve a height is, t = 3.5 s.

Clearly, the ball rising high will need a mathematical formula which utilizes the initial velocity and final velocity (v). So, using the first kinematic equation of motion as,

v = u + gt

Here,

g is the gravitational acceleration.

Solving as,

v = 27 + (9.8 × 3.5)

v = 61.3 m/s

Now, if we calculate the height achieved through the second kinetic equation of motion as,

Thus, we can conclude that the snow ball will achieve the height of 34.75 m and at that height, it will move with a speed of 61.3 m/s.

Learn more about the kinematic equations of motion here:

link

Dy = 111.66 [m]

t  = 3.5 [s]

Explanation:

To solve this problem we must use the equations of kinematics.

where:

Vf = final velocity [m/s]

Vo = initial velocity = 27 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time = 3.5 [s]

Note: The negative sign of the equation means that the gravity acceleration goes in opposite direction

Vf = 27 - (9,81*3,5)

Vf = - 7.33 [m/s] (this negative sign indicates that at this moment the snowball is going downwards)

To find how high the snowball was we must use the following equation:

Dy = (27*3.5) + (0.5*9.81*3.5)

Dy = 94.5 + (17.16)

Dy = 111.66 [m]

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

do you have a formula possibly