Question

Determine the distance R to B for the aunch conditions shown. Ans: R 2386.37m A-150 m/s 150 m

Answer 1

Let us consider the upwards direction as positive and the downwards direction as negative.

Gravitational acceleration = g = -9.81 m/s²

Initial velocity of the projectile = V = 150 m/s

Angle of launch = $\theta$

From the figure,

Sin$\theta$ = 3/5 = 0.6

Cos$\theta$ = 4/5 = 0.8

Initial horizontal velocity of the projectile = V_x = VCos$\theta$ = (150)(0.8) = 120 m/s

Initial vertical velocity of the projectile = V_y = VSin$\theta$ = (150)(0.6) = 90 m/s

Initial height of the point from where it is launched = H = 150 m

Time taken by the projectile to reach the ground = T

When the projectile lands on the ground the displacement of the projectile in the vertical direction is downwards therefore it is negative.

-H = V_yT + gT²/2

-150 = (90)T + (-9.81)T²/2

4.905T² - 90T - 150 = 0

$T = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

$T = \frac{-(-90) \pm \sqrt{(-90)^{2} - 4(4.905)(-150)}}{2(4.905)}$

$T = \frac{90 \pm 105.0856}{9.81}$

T = 19.8864 sec or -1.538 sec

Time cannot be negative.

T = 19.8864 sec

There is no horizontal force on the projectile therefore the horizontal velocity of the projectile remains constant.

Horizontal displacement of the projectile = R

R = V_xT

R = (120)(19.8864)

R = 2386.37 m