a projectile fired at a velocity of 65m/s, 20 degrees above the horizontal lands

Question

a projectile fired at a velocity of 65m/s, 20 degrees above the horizontal lands a certain distance from its launch point. if the launch angle is increased to 30 degrees, what would the new initial speed have to be in order for the projectile to have the same range 5. A projectile fired at a velocity of 65 m/s, 20 above the horizontal lands a certain distance from its launch point. If the launch angle is increased to 30°, what would the new initial speed have to be in order for the projectile to have the same range? (Assume the ground is a horizontal surface.) a) 59.5 m/s b) 57.9 m/s c) 56.0 m/s d) 53.8 m/s e) 52.5 m/s

Explanation / Answer

Horizontal range of a projectile is given by,

R = (V02 * sin (2?))/g

Where V0 is the initial speed, ? is the angle at which projectile is fired

Now V0 is given 65 m/s for the first case and ? is 20 degrees.

For second case, we need to find V, ? is 30 degrees.

For the same range,

(652*sin (2*20))/g = (V2 * sin (2*30))/g

V2 = 652* sin (40)/sin (60)

V2 = 3136

V = 56 m/s

Hence, answer is c) 56 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.