A cart 2 meters in length, moving with a velocity of 4.9 m/sec, is approaching a

Question

A cart 2 meters in length, moving with a velocity of 4.9 m/sec, is approaching a platform which is 9.8 meters above the plane of the cart's motion. The cart is moving along a line marked every 0.1 meters. You want to drop a particle onto the cart

a. If you want to hit the center of the cart, how many seconds before the front of the cart arrives under the platform should you drop the particle?

b. How many meters before the front of the cart arrives?

c. If you want to hit any place on the cart, during what time interval (referenced to the front of the cart) before the cart arrives must you drop the particle?

Explanation / Answer

(A) particle have to drop 9.8 m to reach the truck,

vertical displacement, y = 0 -9.8 = - 9.8 m

v0y = 0

Applying y = v0y t + ay t^2 / 2

- 9.8 = 0 - 9.8 t^2 / 2

t = 1.414 sec

for car: t1 = 1/ 4.9 = 0.204 sec

time before particle was dropped = t - t1 = 1.21 sec ....Ans

(B) d = 4.9 x 1.21 = 5.93 m

(c) t1 = 0 Or

2/4.9 = 0.408 sec

time range = 1 sec to 1.414 sec

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