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 50.0% of its maximum height?
A) 26.0 m/s
B) 27.4 m/s
C) 28.7 m/s
D) 26.7 m/s
E) 28.1 m/s

Question

Quizplus · Accepted Answer

The speed of the projectile is given by:$$v = v_0\sqrt{\frac{2h}{h_{max}}}$$where:* v is the speed of the projectile at height h* v0 is the initial speed of the projectile* h is the height of the projectile* hmax is the maximum height of the projectileWe are given that h = 50.0% of hmax, so:$$h = 0.500h_{max}$$Substituting this into the equation for v, we get:$$v = v_0\sqrt{\frac{2(0.500h_{max})}{h_{max}}}$$$$v = v_0\sqrt{1.00}$$$$v = v_0$$Therefore, the speed of the projectile when it has reached a height equal to 50.0% of its maximum height is equal to its initial speed, which is 30.0 m/s.

A Projectile Is Fired from Ground Level at an Angle