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Question

A projectile is fired at an angle 30 ° from a gun that is 45 m above the flat ground, emerging from the gun with a speed of 300 m/s.

(a) How long does the projectile remain in air?

(b) At what horizontal distance from the firing ground does it strike the ground?

(c) What is the velocity of the projectile when it hits the ground?

Answer 1

A projectile is fired at an angle 30 ° from a gun that is 45 m above the flat ground, emerging from the gun with a speed of 300 m/s.

(a) How long does the projectile remain in air?

(b) At what horizontal distance from the firing ground does it strike the ground?

(c) What is the velocity of the projectile when it hits the ground?

This problem consists of two parts

Let us find T1 as time of normal flight

T1 = 2V Sin theta /g = 2*300* Sin 30 /9.8 = 30.6m/s

T2 = time to cover additional 45 m with initial downward speed of VSin theta = 300*Sin30 = 150 m/s

Solve equation 4.9t² +150t -45 = 0 ..... Eq 1

Get two values of t = T2 and then T = T1 +T2 = total time in air

Total horizontal distance = R1 + R2 = V² Sin 2Theta /g = 300*300 Sin 60 / g +R2 = 7953.3 m +R2

R2 = V CosTheta *t = 300 Cos 30 * t

Plug in t from Eq 1 and get R1 +R2

Velocity when hits the ground = Sqrt ( V x2 + Vy2) = Sqrt (V²Cos² theta + Vy²) = = Sqrt ( 67500 + Vy² )

Vy + Final vertical velocity when it hits the ground

Answer 2

v_x = 300cos30° = 259.8 m/s; v_y = 300sin30° - 10t = 150 - 10t m/s.

x = 259.8t; y = 45 + 150t - 5t².

(a) y = 45 + 150t - 5t² = 0; t² - 30t - 9 = 0; t = 15 ± √234; t = 15 + 15.3 = 30.3 s

(b) x = 259.8·30.3 = 7872.2 m

(c) v = √v_x² + v_y²= √(259.8)² + (150 - 10·30.3)² = 301.5 m/s