Question

A body is projected with speed 40m/s at an angle 37^0with the horizontal from ground.Its time of flight is (g= 10m/s^(2))​

Answer 1

Answer:

u= 40m/s and angle= 37

time of flight = Tf

Tf= 2usin°/g = 2×40×sin37 /10 = 80 × sin37/10= 4.8s

Answer 2

Given:

A body is projected with speed 40m/s at an angle of 37° with horizontal.

To find:

Time of flight ?

Calculation:

The expression of time of flight of an oblique projectile is :

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

$\implies T = \dfrac{2 \times 40 \times \sin( {37}^{ \circ} ) }{g}$

$\implies T = \dfrac{2 \times 40 \times \dfrac{3}{5} }{10}$

$\implies T = 2 \times 4 \times \dfrac{3}{5}$

$\implies T = 4.8 \: sec$

So, time of flight is 4.8 sec.