Question

A ball is thrown at an angle theta with the horizontal. Its horizontal range is 60 m and time of flight is 3 s. What is the horizontal component of its velocity of projection?

Answer 1

Step-by-step explanation:

this is your answer. ..

Answer 2

Answer:

The horizontal component of its velocity of projection  $= 20 \ \frac{m}{s}$.

Step-by-step explanation:

As per the question a ball is thrown at an angle of theta with horizontal.

Given data are:

Horizontal range = 60 , which is denoted as S

∴ S = 60

Time of flight (t) = 3 s

As the acceleration here be = a =0

Now suppose the horizontal component of its velocity of projection is ''u''.

Using this formula,

$S = ut + \frac{1}{2} at^{2}$

$60 = u \times 3 +\frac{1}{2} \times 0 \times 3^{2}$

$u=20 \ \frac{m}{s}$

Hence,  the horizontal component of its velocity of projection  $= 20 \ \frac{m}{s}$