Question

The equation of motion of a projectile are given by x=36t meter and 2y=96t - 9.8t^2 meter. The angle of projection is

Answer 1

Answer: $\theta = sin^{-1}\dfrac{4}{5}$

Explanation:

Given that,

$x = 36 t$...(I)

$2y = 96t-9.8t^{2}$...(II)

Using the kinematic equation of displacement of projection of motion

$x = ut\cos\theta$...(III)

$y = ut\sin\theta - \dfrac{1}{2}gt^{2}$...(IV)

On Comparing equations(I), (III) and (II), (IV)

$u\cos \theta = 36$.....(A)

$u\sin\theta = 48$.....(B)

Now, the angle of projection is

Divided equation (B) by equation (A)

$\dfrac{u\sin\theta}{u\cos\theta} = \dfrac{48}{36}$

$tan\theta = \dfrac{4}{3}$

So, $sin\theta = \dfrac{4}{5}$

$\theta = sin^{-1}\dfrac{4}{5}$

Hence, the angle of projection is $\theta = sin^{-1}\dfrac{4}{5}$.

Answer 2

According to the data given, the equations are $x = 36 t$ and $2 y = 96 t - 9.8 t^2$

Thereby, as we know the component of velocity of a projectile we can compare the equation.

As $x = 36 t$ can be compared to $u\cos { \theta }$

Therefore, from here, we have $\theta = 36$ and

similarly, $2 y = 96 t - 9.8 t ^2$

$\Rightarrow y = 48t - 4.9 t^2$, compared with $y = u sin\theta (t) - \frac {1}{2} t^2$

From the above formula, we have $u sin \theta= 48$

Therefore $\frac {u sin \theta}{u cos\theta} = \frac {48 }{36}$

$\Rightarrow tan \theta = \frac {4}{3}$

$\Rightarrow sin\theta = \frac {4}{5}$

$\Rightarrow \theta =\sin ^{ -1 }{ \frac{ 4 }{ 5 } }$

$\Rightarrow \theta=53 \quad degree$