A. toy car is released from rest on a ramp h_1=30cm above the ground, moving ont

Question

A. toy car is released from rest on a ramp h_1=30cm above the ground, moving onto a launch ramp which reaches to a maximal height of h_2=l0cm above the ground (neglect friction and the rotational energy of the wheels). What is the speed of the car once it becomes airborne? If the car reaches a maximal height of 15cm above the ground during its flight, what is the inclination angle theta of the launch ramp with respect to the horizontal? How long is the car in the air? What is the horizontal flight distance?

Explanation / Answer

a) By energy conservation,

PEf +KEf = PEi + KEi

m*9.8*0.1 + 0.5*m*v^2 = m*9.8*0.3 + 0

0.98 + 0.5v^2 = 2.94

v^2 = 2*1.96 =3.92

v = 1.98 m/s

b) Vy = sqrt (2gh) = sqrt (2*9.8*(0.15-0.1)) = 0.99 m/s

sin theta = 0.99/1.98 = 1/2

theta = arcsin 0.5 = 30 degree

c) The launch point is 0.1 m high, and after striking the ground the displacement from launch point is -0.1m

s = ut - 0.5gt^2

   -0.1 = 0.99 t - 0.5*9.8 t^2

4.9 t^2 -0.99t -0.1 =0

t = 0.276 s

d) horizontal flight distance = v cos 30 degree t

       = 1.98 * 0.866*0.276

      =0.473 m = 47.3 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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