A projectile is launched with an initial speed of 58.0 m/s at an angle of 34.0 a

Question

A projectile is launched with an initial speed of 58.0 m/s at an angle of 34.0 above the horlzontal. The projectile lands on a hillside 3.95 s later. Neglect air friction. (Assume that the +x-axis is to the right and the y-axis is up along the page.) (a) What is the projectile's velocity at the highest point of its trajectory? magnitude 48.1 direction o m/s counterclockwise from the + x-axis b) What is the straight-line distance from where the projectile was launched to where it hits its target? 115 Your response differs from the correct answer by more than 10%. Double check your calculations m Need Help? Read N

Explanation / Answer

a) The velocity at its highest point will be having x component only which is constant throughout the motion and equal to 58 cos(34 deg) = 48.08 m/s
angle = 0
b) vertical velocity = 58 sin(34 deg) = 32.43 m/s
horizontal speed = 48.08 m/s
so vertical height = 32.43*3.95-0.5*9.81*3.95^2 = 51.568 m
horizontal displacement = 32.43*3.95 = 128.098 m
so the net distance = 138 m.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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