Question

A body projected by making an angle 30 degree with horizontal with a velocity of 39.2 MS -1 find Time of flight, Range and Maximum height??​

Answer 1

Answer:

T = 2usinO/g

R = u^2 sin2O/g

H = u^2 sin^2O/2g

O - Thita - Angle of Protection

in Your Question O - 30°

U - velocity - 39.2m/s

g - Acceleration due to Gravity -9.8m/S^2

Put Value Get answer

Answer 2

The Time of flight, Range and Maximum height are 4 seconds, 135.79 meters and 19.6 meters respectively.

Explanation:

Given that,

Initial speed of the projectile, u = 39.2 m/s

Angle of projection, $\theta=30^{\circ}$

To find,

Time of flight, Range and Maximum height.

Solution,

Time of flight is given by :

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

$T=\dfrac{2\times 39.2\ sin(30)}{9.8}$

T = 4 seconds

Range of projectile is given by :

$R=\dfrac{u^2\ sin2\theta}{g}$

$R=\dfrac{(39.2)^2\ sin2(30)}{9.8}$

R = 135.79 meters

Maximum height of a projectile is given by :

$H=\dfrac{u^2sin^2\theta}{2g}$

$H=\dfrac{(39.2)^2sin^2(30)}{2\times 9.8}$

H = 19.6 meters

So, the Time of flight, Range and Maximum height are 4 seconds, 135.79 meters and 19.6 meters respectively.

Learn more,

Projectile motion

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