A body thrown vertically upward returns to the earth in three seconds.(a)what was the initial velocity of the body (b)what height did the body reach? Neglect air resistance.

1 Answer
Aug 31, 2017

a 14.7 m/s
b 11 meters


A couple of facts we know. Gravitational acceleration in the vicinity of the earth's surface can be treated as a constant 9.81 m/s.

This is #(d^2x)/dt^2#

With this, we can write a function for velocity v as a funciton of t:

#v(t) = dx/dt = int-9.81dt = v_0 - 9.81t#

(We are assigning a NEGATIVE value to downward acceleration & velocity)

We are given total flight time: 3 seconds.

So we know that at the halfway point (1.5 seconds), the ball will have reached maximum height, and v at that time will be 0.

The ball then accelerates downwards for the remaining 1.5 seconds.

Over this time it accelerates to #-9.81 * 1.5 = -14.7 m/s#.

Since we're ignoring wind resistance, the downwards velocity at time t = 3 is equal in magnitude (opposite in direction) to the initial velocity. So this is your answer to part a: 14.7 m/s.

(I'm rounding to 3 significant digits in all arithmetic calculations).

We can then write an equation for the height, x, as a function of t:

#x(t) = int(v_0 - 9.81t)dt = (v_0* t) -(9.81/2)t^2 + c#

At time t = 0, height x = 0, so our constant c is also 0.

Plugging in time t = 1.5, we can solve for the height at that time:

#x(1.5) = 14.7(1.5) - 4.91(1.5^2) = 11 meters #

(once again, rounding to 3 significant digits.)