A ball is thrown vertically upwards with an initial velocity of 27 m per sec Neglecting air resistance how long is the ball in the air?

Answer 1

For this question recall the SUVAT formula, and the symettry argument for a projectile undergoing no resistance.

Use V=U + A x t (eqn 1)

Where

U is initial velocity

V is final velocity

A is acceleration

t is the time taken

Now using symettry arguments, U=-V=27m (taking upwards as positive)

Now, depending on the value of acceleration (solely gravity in this case), we can use A= -10, -9.8 or even -9.81 (negative because it is acting downward- this is key to the use of SUVAT formula without vectors) as standard. I will solve the problem with the first two, but show working so that you can input your own numbers if you wish:

rearrranging eqn 1:

V=U+At

V-U=At

t=(V-U)/A

For A=-10N/kg (m/s2)

t=(-27-27)/-10

t=54/10

t=5.4 seconds

For A=-9.8N/kg (m/s2)

t=-54/-9.8

t=5.5102 seconds (4 d.p.)

Oh and just in case you were interested:

Using the formula S=Ut+0.5at2 you can work out the total distance travelled:

S=27x5.4+0.5x-10x5.42

S=0

This result occurs because the total distance travelled is nothing, as the ball is thrown straight up and it returns to the same place.

To find the maximum height, using symmetry, halve the time:

S=27x2.7+0.5x-10x2.72

S=36.45 m (this is the maximum height)