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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

User Fechnert
by
8.1k points

2 Answers

6 votes

Answer:


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

A is correct.

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

User Ostrichofevil
by
8.5k points
5 votes

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

User Isomarcte
by
8.4k points
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