The labeled graphs below represent the upward trajectories of 4 bodies, two slid

Question

The labeled graphs below represent the upward trajectories of 4 bodies, two sliding upwards on frictionless inclines and two in free flight. Note that all 4 bodies reach the same maximum height (9 meters) after traveling from the same initial elevation (0 m) 10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Relative horizontal position, x, (m) The time of travel of P is equal to that of Q The time of travel of R is ual othat ofN The initial speed of Q is less thanthat of P The time of travel of N is greater thanthat of Q The initial speed of N is less thanthat of R The initial speed of Q is grester thanthat of N Add hint text here. Submit Answer Tries 2/12 Previous Tries

Explanation / Answer

(a) The time of travel of P is GREATER THAN that of Q.

Explanation -

Horiontal displacement of P > horiontal displacement of Q.

so P will take more time than Q.

(b) The time of travel of R is GREATER THAN of N.

Explanation -

Same reason as mentioned for (a).

(c) for P -

(V1*sin theta1)^2 / 2g = 9

=> V1*sin theta1 = sqrt(18*g) = 13.3 = V2*sin theta 2(for Q)

again -

V1*cos theta1 * t1 > V2*cos theta2 * t2

from these two we can conclude -

The initial speed of Q is LESS THAN that of P. (your option is correct)

(d) The time of travel of N is LESS THAN that of P.

(e) The initial speed of N is LESS THAN that of R.

(f) The initial speed of Q is EQUAL TO that of N.

Explanation -

Because for the two paths, horiontal and vertical displacements are the same.

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