Question

Projectiles P 60° 50 m s -1 An object is projected from P with velocity 50 m s'inclined at 60° to the horizontal PQ. Find: (a) time of flight (b) range PQ (c) greatest height above PQ (d) height above PQ after 2 seconds (e) direction of flight after 3 seconds.​

Answer 1

a) The time of flight is approximately 5.102 seconds. b) The range PQ is approximately 176.48 meters. c) The greatest height above PQ is approximately 63.98 meters.

Step-by-step explanation:

a) To find the time of flight, we can use the formula t = 2v*sin(theta)/g, where v is the initial velocity and theta is the angle of projection. Plugging in the values, we get t = 2 * 50 * sin(60°) / 9.8 ≈ 5.102 seconds.

b) The range PQ can be calculated using the formula R = v²*sin(2*theta)/g. Plugging in the values, we get R = 50²*sin(120°)/9.8 ≈ 176.48 meters.

c) The greatest height above PQ can be found by using the formula H = (v²*sin²(theta))/(2*g). Plugging in the values, we get H = (50²*sin²(60°))/(2*9.8) ≈ 63.98 meters.

d) To find the height above PQ after 2 seconds, we can use the formula h = v*sin(theta)*t - (1/2)*g*t². Plugging in the values, we get h = 50*sin(60°)*2 - (1/2)*9.8*2² ≈ 56.72 meters.

e) The direction of flight after 3 seconds can be determined by finding the horizontal and vertical components of velocity at that time. The horizontal component remains constant, so it will be 50*cos(60°). The vertical component can be found using the formula v_y = v*sin(theta) - g*t, where t is the time. Plugging in the values, we get v_y = 50*sin(60°) - 9.8*3 ≈ 10.435 m/s. Therefore, the direction of flight after 3 seconds is given by the angle tan^(-1)(10.435/50*cos(60°)).