Question

8. At a Punkin' Chunkin' contest a pumpkin is launched at an angle of 65° to the horizontal with a speed of 15 m/s. Determine (5+5+5=15 pts) (a) Range and time of flight of the pumpkin. (b) The position vector of the pumpkin at t = 0.5 s assuming the launch location to be the origin.

          8. At a Punkin' Chunkin' contest a pumpkin is launched at an angle of 65° to the
horizontal with a speed of 15 m/s. Determine (5+5+5=15 pts)
(a) Range and time of flight of the pumpkin.
(b) The position vector of the pumpkin at t = 0.5 s assuming the launch location to
be the origin.

8. At a Punkin' Chunkin' contest a pumpkin is launched at an angle of 65° to the
horizontal with a speed of 15 m/s. Determine (5+5+5=15 pts)
(a) Range and time of flight of the pumpkin.
(b) The position vector of the pumpkin at t = 0.5 s assuming the launch location to
be the origin.

Added by Manuela L.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Rahul Kumar

10/18/2022

Step 1

The horizontal component of the initial velocity (Vx) can be found using the equation Vx = V * cos(θ), where V is the speed and θ is the launch angle. Vx = 15 m/s * cos(65°) = 15 m/s * 0.4226 ≈ 6.34 m/s The vertical component of the initial velocity (Vy) can be Show more…

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8. At a Punkin' Chunkin' contest a pumpkin is launched at an angle of 65 to the horizontal with a speed of 15 m/s. Determine (5+5+5=15 pts) (a) Range and time of flight of the pumpkin. (b) The position vector of the pumpkin at t = 0.5 s assuming the launch location to be the origin.