Question

A particle is projected up inclined plane such that its component of velocity along tge incline is 10m/s. Time of flight is 2sec & maximum height above incline is 5m.Find the velocity of projection?

Answer 1

Derive the required formulas for motion up the inclined plane and solve the question. I solved it as follows (see attachment)

Hope it helps!!

Answer 2

Answer:  

The initial velocity of the projectile is $\bold{10 \sqrt{2}\ \mathrm{m} / \mathrm{s}}$.

Solution:

The given quantities are initial velocity along the incline i.e., $u \cos \theta=10\ \mathrm{m} / \mathrm{s}.$

Time of flight t = 2s

Maximum height H = 5m.

We know that the velocity of projection:

$u=u \cos \theta i+u \sin \theta j \rightarrow(1)$

So we have to determine the value of u sinθ, in order to find the velocity of projection.

The time of flight can be found using the formula

$\begin{array}{l}{t=\frac{2 u \sin \theta}{g \cos \alpha}} \\ \\ {2=\frac{2 u \sin \theta}{10 \cos \alpha}}\end{array}$

Since, g is acceleration due to gravity which is considered as $10\ \mathrm{m} / \mathrm{s}^{2}$ in this case. Here α is the angle formed between the direction of projection and ground

So we get,

$10 \cos \alpha=u \sin \theta \rightarrow(2)$

From the Height of projection:

$H=\frac{u^{2} \sin ^{2} \theta}{2 g \cos \alpha}$

$5=\frac{u^{2} \sin ^{2} \theta}{20 \cos \alpha}$

$100 \cos \alpha=u^{2} \sin ^{2} \theta \rightarrow(3)$

We can get the value for u sin θ by dividing equation (3) with equation (2), we get

$\frac{100 \cos \alpha}{10 \cos \alpha}=\frac{u^{2} \sin ^{2} \theta}{u \sin \theta}$

$10=u \sin \theta \rightarrow(4)$

Substitute (4) in (1)

$u=10 i+10 j$

$|u|=\sqrt{(10)^{2}+(10)^{2}}=10 \sqrt{2}\ \mathrm{m} / \mathrm{s}$

Thus, the initial velocity of the projectile is $\bold{10 \sqrt{2}\ \mathrm{m} / \mathrm{s}}$.