Question

If a ball is thrown into the air with a velocity of 34 ft/s, its height (in feet) after t seconds is given by y = 34t − 16t2. Find the velocity when t = 1.

Answer 1

ANSWER:

If a ball is thrown into the air with a velocity of 34 feet per second, then velocity of the ball after 1 second is 2 feet per second

SOLUTION:

Given, a ball is thrown into the air with a velocity of 34 feet per second

Initial velocity (u) = 34 feet per second

And also given a relation between displacement and time = $\mathrm{y}=34 \mathrm{t}-16 \mathrm{t}^{2}$ --- eqn 1

We need to find the velocity when t = 1 ; v = ?

We know that, v = u + at and $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

where v is instantaneous velocity and u is initial velocity

a is acceleration

t is time interval  

s is displacement

using the displacement and time relation eqn (1) we get

Now, when t = 1, displacement s = 34(1) – 16(1)

$\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}=34-16$

$34 \times 1+\frac{1}{2} \times a \times 1^{2}=18$

$34+\frac{a}{2}=18$

$\begin{array}{l}{\frac{a}{2}=18-34} \\\\ {\frac{a}{2}=-16} \\ {a=-16 \times 2} \\ {a=-32}\end{array}$

here, -ve sign indicates that object is in deceleration . so acceleration is -32 ft/s

now put a value in v = u + at

v = 34 + (-32)(1)

v = 34 – 32

v = 2 ft/s

Hence, velocity of the ball after 1 second is 2 ft/s