A projectile is fired from ground level at an angle of 40.0° above horizontal at a speed of 30.0 m/s. What is the speed of the projectile when it has reached a height equal to 0.500 of its maximum height?

Question

Quizplus · Accepted Answer

Explanation:At the highest point, the vertical component of velocity is zero. So, the speed at the highest point is equal to the horizontal component of velocity. The horizontal component of velocity is constant throughout the projectile's motion. So, the speed at the highest point is equal to the initial speed times the cosine of the launch angle.At a height equal to 0.500 of its maximum height, the vertical component of velocity is equal to the negative of the vertical component of velocity at the highest point. So, the speed at a height equal to 0.500 of its maximum height is equal to the square root of the sum of the squares of the horizontal component of velocity and the negative of the vertical component of velocity at the highest point.The horizontal component of velocity is equal to the initial speed times the cosine of the launch angle. The vertical component of velocity at the highest point is equal to the initial speed times the sine of the launch angle. So, the speed at a height equal to 0.500 of its maximum height is equal to:$$v = \sqrt{(v_i \cos 	heta)^2 + (-v_i \sin 	heta)^2}$$$$v = v_i \sqrt{\cos^2 	heta + \sin^2 	heta}$$$$v = v_i$$So, the speed at a height equal to 0.500 of its maximum height is equal to the initial speed. The initial speed is 30.0 m/s. So, the speed at a height equal to 0.500 of its maximum height is 30.0 m/s.

A Projectile Is Fired from Ground Level at an Angle