Question

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?

Answer 1

Answer:

Dy = 111.66 [m]

t  = 3.5 [s]

Explanation:

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

$v_{f} = v_{o} - (g*t)$

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:

$Dy=v_{o} *t+\frac{1}{2}*g*t$

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

Dy = 94.5 + (17.16)

Dy = 111.66 [m]