Question

A stone is thrown at an angle of 30° from the ground with an initial velocity of 60
m/s. It falls on the ground at a distance d from the initial point.
Take g = 10 m/s?.
(a) What is the time taken by the stone to reach the ground?

(b) What is the value of d?
m
(c) What is the maximum height reached by the stone? ​

Answer 1

(i) Time taken = 2usin theta /g

Here theta = 30°

And u= 60 m/s

Putting the values in the formulas

We get T= 2×60×1/2g

T= 6 seconds.

(ii) here d is the range of the projectile

And d= u^2 sin 2theta/g

D = 180√3 m

(iii) maximum height.

= u^2 sin^2 theta/2g

= 45m.

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Answer 2

We're given,

$\theta=30\textdegree\quad;\quad\sin\theta=\dfrac{1}{2}\quad;\quad\cos\theta=\dfrac{\sqrt3}{2}\\\\u=60\ ms^{-1}$

Here stone can be considered as a projectile. So  $d$  is the horizontal range of the projectile.

(a)  We have,

$t=\dfrac{2u\sin\theta}{g}=\dfrac{2\cdot60\cdot\dfrac{1}{2}}{10}=\mathbf{6\ s}$

(b)  We have,

$d=\dfrac{u^2\sin(2\theta)}{g}=\dfrac{60^2\sin60\textdegree}{10}=360\cdot\dfrac{\sqrt3}{2}=\mathbf{180\sqrt3\ m}$

(c)  We have,

$H=\dfrac{u^2\sin^2\theta}{2g}=\dfrac{60^2\cdot\dfrac{1}{4}}{20}=\mathbf{45\ m}$