Question

20. A ball is thrown with a speed of 17 ms-1 at a pro
jection angle of 53 degree ii) the maximum height above the release point.​

Answer 1

Given :

$\large\rm {u = 17 ms^{-1} }$

$\large\rm { θ = 53° }$

$\large\rm { g = 10 \ ms^{2}}$

We get $\large\rm { sin \ θ = sin \ 53° = \frac {4}{5}}$

Maximum height reached:-

$\large\rm { H = \frac {u^{2} \ sin^{2} θ}{2g}}$

$\large\rm {= \frac {17 × 17 × ( \frac {1}{2} )^{2}}{20}}$

So,

$\large\rm { H = 3.6125}$