Question

A ball is thrown with 5m/s at angle of projection 37 degrees . Find time of flight, maximum height , range?​

Answer 1

Explanation:

time of flight=5xsin37⁰/g(10)

max height=5²x sin²37⁰/2g

Range=5xsin2x37⁰/g

just put the values and solve bro

Answer 2

Given:

A ball is thrown with 5m/s at angle of projection 37°.

To find:

• Time of flight
• Maximum height
• Range

Calculation:

• Please remember the following FORMULAS for projectiles.

$1) \: t = \dfrac{2u \sin( \theta) }{g}$

$\implies \: t = \dfrac{2 \times 5 \sin( {37}^{ \circ} ) }{10}$

$\implies \: t = \dfrac{2 \times 5 \times \frac{3}{5} }{10}$

$\implies \: t = 0.6 \: sec$

So, time of projectile is 0.6 seconds.

Now, max height be h :

$h = \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g}$

$\implies h = \dfrac{ {5}^{2} \times { \sin}^{2}( {37}^{ \circ} ) }{2 \times 10}$

$\implies h = \dfrac{25 \times \frac{9}{25} }{2 \times 10}$

$\implies h = \dfrac{9}{20}$

$\implies h =0.45 \: m$

So, max height is 0.45 metres.

Now, let range be R :

$R = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$\implies R = \dfrac{ {5}^{2} \times 2\sin( \theta) \cos( \theta) }{10}$

$\implies R = \dfrac{ {5}^{2} \times 2\sin( {37}^{ \circ} ) \cos( {37}^{ \circ} ) }{10}$

$\implies R = \dfrac{ {5}^{2} \times 2 \times \frac{3}{5} \times \frac{4}{5} }{10}$

$\implies R = \dfrac{24 }{10}$

$\implies R = 2.4 \: m$

So, range is 2.4 metres.