Ok, so I know this is really late, but for the future students who will be stuck like me, here how I did it with Range Formula. Range Formula gives you the distance of X.

(V^2)(sin2*degree given)

R=

gravity

You are given Distance of x, which is 12m. So R=12, gravity will be 9.8 for these sorts of problem, so its like a constant variable. Plugging it in, you get:

V^2sin(2*15)

12=

9.8

Let:

v be his speed,

t be the time from take-off,

a be the angle of the incline to the horizontal,

g be the acceleration due to gravity,

x be the horizontal distance travelled,

s be the width of the canyon.

Resolving horizontally and vertically:

x = vt cos(a) (1)

0 = vt sin(a) gt^2 / 2 (2)

x >= s (3)

From (1):

t = x / v cos(a)

Substituting this value of t in (2):

0 = x sin(a) / cos(a) gx^2 / 2v^2 cos^2(a)

Multiplying by 2v^2 cos^2(a):

0 = 2xv^2 sin(a)cos(a) gx^2

x (v^2 sin(2a) gx) = 0

x = 0 (initially)

or

v^2 = gx / sin(2a)

Using (3):

v^2 >= gs / sin(2a)

v >= sqrt(9.81 * 12 / sin(30))

= sqrt(9.81 * 12 * 2)

= 15.34 m/s.

On reaching the other side, his horizontal velocity will be unchanged, as gravity acts only vertically.

His vertical velocity will be the same magnitude as it was on take off, but downward instead of upward.

Combining these two components, his speed on landing will be 15.34 m/s.

Ok, so I know this is really late, but for the future students who will be stuck like me, here how I did it with Range Formula. Range Formula gives you the distance of X.

(V^2)(sin2*degree given)

R=

gravity

You are given Distance of x, which is 12m. So R=12, gravity will be 9.8 for these sorts of problem, so its like a constant variable. Plugging it in, you get:

V^2sin(2*15)

12=

9.8

12(9.8)

= V^2

sin(30)

V= 15.336 m/s

Which is your resultant vector and what you need as a min.

This post is last updated on hrtanswers.com at Date : 1st of September – 2022