Question

I Use a (t) = — 32 f_2 as the acceleration due to gravity. A ball is thrown vertically upward from the S ground with an initial velocity of 56 feet per second. For how many seconds will the ball be going upward? Express your answer as a decimal rounded to 1 decimal place.

Answer 1

Explanation:

It seems there is an error in the expression you provided for the acceleration due to gravity. The correct expression for the acceleration due to gravity is "a(t) = -32 ft/s²" (assuming you meant the acceleration due to gravity on Earth). The negative sign indicates that the acceleration is directed downward.

Now, let's proceed with the problem based on this corrected expression.

Given:

Acceleration due to gravity, a(t) = -32 ft/s²

Initial velocity, u = 56 ft/s (thrown vertically upward)

Position at time t, s(t) = ?

To find the position at any given time, we can integrate the equation for velocity with respect to time, and then integrate the equation for position with respect to time.

The equation for velocity is:

v(t) = u + ∫(a(t)) dt

Substituting the value of a(t):

v(t) = u + ∫(-32) dt

v(t) = u - 32t + C₁ (where C₁ is the constant of integration)

To find the constant of integration, we need additional information. If the ball starts at the ground (s = 0) when t = 0, then the initial position can be expressed as s(0) = 0.

The equation for position is:

s(t) = ∫(v(t)) dt

Substituting the expression for v(t):

s(t) = ∫(u - 32t + C₁) dt

s(t) = ut - 16t² + C₁t + C₂ (where C₂ is the constant of integration)

Again, we can determine the constant of integration, C₂, using the initial condition s(0) = 0:

s(0) = 0 = 0 - 16(0)² + C₁(0) + C₂

C₂ = 0

Therefore, the equation for the position of the ball at any given time t is:

s(t) = ut - 16t² + C₁t

Now, if you provide the time t for which you want to calculate the position, I can help you determine the value of s(t).