A ball is thrown vertically upward from the ground with an initial velocity of 103ft/sec. Use h=−16t2+v0t+h0.

Question

Can you explain every step

A ball is thrown vertically upward from the ground with an initial velocity of 103ft/sec103⁢ft/sec. Use h=−16t2+v0t+h0h=−16t2+v0t+h0.

Answer 1

Hi Mike, my name's Mike, too.

h = -16t² +v₀t + h₀ is an extremely important formula. (On Earth, anyway)

It describes the height (h) of a free-falling object as a function of time.

The ONLY variable is "t" and it's measured in seconds.

The h₀ and v₀ are not variables, but constants.

Constants change as you move from one problem to another, but NEVER change within a problem.

The subscripted 0's are use to indicate initial conditions, v₀ is the velocity and h₀ is the height when t = 0.

In some books you might see v_i and h_i, where the "i" means initial.

Notice the -16 is a constant that does not change because it's associated with Earth's gravity.

It should be noted that if you measure the height in meters instead of feet, the constant would be -4.9.

Notice, that both the numbers are negative, this is because gravity is directing the object downward.

You would need a different constant for each unit of length you might decide to use.

You would definitely need a different constant if you moved to another planet.

In this problem, we're told that the initial velocity (v₀) is 103 ft/sec.

The positive indicates that it's in the upward direction.

(If you threw an object straight DOWN from a cliff, the v₀ would be negative and the h₀ would be the height of the cliff.)

Since this ball was initially on the ground, v₀ = 0.

So based upon initial conditions the height equation is; h = -16t² + 103t.

I hope this helps.

And if you ever get the chance to play ball on another planet, please let me come with you.