Question

An objoct is dropped from a tower. 159 ft above the ground. The object's height above ground t sec into the fall is 5=159−16t 2

a. What is the object's velocity, speed, and acceleration at time t?
b. About how long does it take the object to hit the ground?
c. What is the object's velocity at the moment of impact? 1 The object's velocity at time t is 2
a The object's speed at time t is 3
b The object's acceleration at time t is ft/sec 2

Answer 1

The object's velocity at time t is -32t ft/sec, speed is 32t ft/sec, and acceleration is -32 ft/sec^2.

Step-by-step explanation:

To find the object's velocity at time t, we need to first find its acceleration. The equation for the object's height above ground at time t is given by 5 = 159 - 16t^2. Taking the derivative with respect to time, we find that the acceleration is -32 ft/sec^2. The velocity is then found by integrating the acceleration with respect to time, and the initial condition is given by v(0) = 0. So the object's velocity at time t is v(t) = -32t ft/sec.

The object's speed at time t is the magnitude of its velocity, so the speed is given by |v(t)| = |-32t| = 32t ft/sec.

The object's acceleration at time t is given by a(t) = -32 ft/sec^2.