A ball is launched straight up into the air from the ground with an initial velocity of 80 feet per second. How long will the ball be in the air?

Question

A ball is launched straight up into the air from the ground with an initial velocity of 80 feet per second. How long will the ball be in the air?

Answer 1

The ball starts with initial velocity v_i = 80 ft/s upwards, rises to the top where its speed is momentarily zero, and then drops back down. Right before it hits the ground it will have final velocity v_f = -80 ft/s. The negative sign of v_f means it points downwards.

The projectile motion is always symmetric: the vertical velocities at the same height on the upward and downward motion have the same magnitudes but opposite directions.

Solve the velocity formula for the time:

v_f = v_i + gt

t = (v_f - v_i) / g = ((-80 ft/s) - 80 ft/s) / (-32 ft/s²) = 5 seconds.

Note: the gravitational acceleration g = -32 ft/s² is a vector and its vertical component is negative because it is pointing downwards, opposite of the chosen upward positive direction of the y axis.

Answer 2

h = -16t^2 +80t

h' = -32t +80 = 0

t= 80/32 = 20/8= 5/2 = 2.5

2.5 x 2 = 5 seconds in the air

It reaches max height in 2 1/2 seconds

then reaches ground in twice that time = 5 seconds

that's a calculus method

or do it algebraically Rewrite the height equation in vertex form, complete the square

h = -16t^2 +80t = -16(t^2 +5t) = -16(t^2 +5t + (5/2)^2) + 16(5/2)^2

= -16(t-5/2)^2 + 16(25)/4 = -16(t-5/2)^2 + 100

where a(x-h)^2 + k is the general form with (h,k) = vertex = maximum = (5/2,100)

5/2 = time to reach the maximum height = 100 feet

double 5/2 to get 5 seconds to reach 0 height