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 $g \approx 32$ 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 $\theta = 45 ^ { \circ }$ ? If necessary, round the answer to the nearest tenth of a second.
A) $1.3 \mathrm { sec }$
B) $0.3 \mathrm { sec }$
C) $1.4 \mathrm { sec }$
D) $1.1 \mathrm { sec }$

Question

Quizplus · Accepted Answer

The correct answer is A.Explanation:Substitute $a = 14$, $\theta = 45 ^ { \circ }$, and $g \approx 32$ into the formula. Then simplify.$$t = \sqrt { \frac { 2 (14) } { 32 \sin 45 ^ { \circ } \cos 45 ^ { \circ } } } = \sqrt { \frac { 28 } { 32 ( \frac { \sqrt { 2 } } { 2 } ) ( \frac { \sqrt { 2 } } { 2 } ) } } = \sqrt { \frac { 28 } { 32 ( \frac { 1 } { 2 } ) ( \frac { 1 } { 2 } ) } } = \sqrt { \frac { 28 } { 8 } } = \sqrt { 3.5 } \approx 1.3 \mathrm { sec }$$

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

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