Solve the problem.
-If friction is ignored, the time t (in seconds) required for a block to slide down an inclined plane is given by the formula $t = \sqrt { \frac { 2 a } { g \sin \theta \cos \theta } }$ where a is the length (in feet) of the base and g =32 feet per second per second is the acceleration of gravity. How long does it take a block to slide down an inclined plane with base a = 14 when θ = 30°? If necessary, round the
Answer to the nearest tenth of a second.
A) 0.4 sec
B) 2.4 sec
C) 1.4 sec
D) 1.1 sec

Question

Quizplus · Accepted Answer

To solve for the time $t$, substitute $a = 14$ feet, $g = 32$ feet/second^2, and $\theta = 30°$ into the formula $t = \sqrt { \frac { 2 a } { g \sin \theta \cos \theta } }$. Using the trigonometric identities for $\sin 30° = 0.5$ and $\cos 30° = \sqrt{3}/2$, the calculation yields $t \approx 1.4$ seconds.

Solve the Problem t=2agsin⁡θcos⁡θt = \sqrt { \frac { 2 a } { g \sin \theta \cos \theta } }t=gsinθcosθ2a​​

Solve the Problem $t = \sqrt { \frac { 2 a } { g \sin \theta \cos \theta } }$