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Time of flight = 2u sinФ/g

Maximum height = (u sinФ)^2 / 2g

Maximum height = (u sinФ)^2 / 2g

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

Total Time of flight = T =2 * (u sin Ф) / g

H = (ut -1/2 a t² formula)

= u Sin Ф * (u sin Ф / g) - 1/2 g (u SinФ/g)²

= u² Sin² Ф / 2 g

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I am deriving the formulas for you here in detailed manner.

Let us say the the projectile is fired into air at an angle of

__Ф__with the horizontal and with an initial speed of

__u__in that direction. Initially, at time t = 0sec, the projectile is at origin O(0,0) at height =0. At time t sec, the projectile is on the parabolic path at a point P (x, y) with instantaneous velocities Vx, and Vy in x and y directions.

There is a deceleration in the vertical direction = - g

Speed of projectile in the vertical direction is :

V = u + a * t => Vy = u sin Ф - g t -- equation 1

Speed in the horizontal direction is a constant as there is no acceleration:

=> Vx = u cos Ф + 0 * t -- equation 2

Displacement = Distance traveled is : s = u t + 1/2 a t²

=> Sx = x = (u cos Ф) t --- equation 3

t = x / (u cos Ф)

Sy = y = (u Sin Ф) t - 1/2 g t²

y = (u sin Ф) (x / u cos Ф) - 1/2 g (x² / u² Cos² Ф)

y = x tan Ф - g x² / (2u²Cos²Ф) --- equation 4

This is the equation of motion of a 2-d projectile and its locus.

====================================

The maximum height H is reached by projectile when Vy becomes 0.

=> Vy = 0 = u sin Ф - g t

=> t = u sin Ф / g = time interval to reach the maximum height.

=> H = u Sin Ф * (u sin Ф / g) - 1/2 g (u SinФ/g)²

=> H = u² Sin² Ф / 2 g -- equation 5

Time of flight = 2 * (u sin Ф) / g --- equation 6

y = 0 => x = R = Range of the projectile , where it lands on ground.

Substituting in equation 4, we get

R = u² Sin 2Ф / g -- equation 6