Question

Mei throws a snowball from a starting height of 10 feet, with an initial velocity of 12 feet per second. Which of the following equations shows the path of the snowball?

H(t) = −16t2 + 12t + 10
H(t) = −16t2 + 10t + 12
H(t) = −16t2 + 22t
H(t) = −16t2 + 22

Answer 1

The form of the equation for ballistic motion with units in feet and seconds is

$H(t)=-16t^2+v_{0}t+h_{0}$

Your problem statement tells you the initial velocity v₀ is 12 ft/s and the initial height h₀ is 10 ft. Substituting these values into the form above gives

... H(t) = -16t² +12t +10 . . . . . . corresponds to the 1st selection

Answer 2

Answer:

$h(t)=-16t^2+12t+10$

A is correct.

Step-by-step explanation:

Mei throws a snowball from a starting height of 10 feet.

We have motion equation if throws up

$h(t)=-16t^2+v_0t+h_0$

where,

$V_0\rightarrow \text{ Initial velocity}=12\ ft/s$

$h_0\rightarrow \text{ Initial Height}=10\ feet$

We need to find the the path of snowball whose initial height 10 feet and initial velocity is 12 feet per second.

Substitute h and v into formula

$h(t)=-16t^2+12t+10$

Hence, The path of snowball is $h(t)=-16t^2+12t+10$