Question

Problem 3.29 King Arthur?s knights use a catapult to launch a rock from their vantage point on top of the castle wall, 13m above the moat. The rock is launched at a speed of 25m/s and an angle of 4O degree above the horizontal.Part A How far from the castle wall does the launched rock hit the ground? Express your answer using two significant figures. D=

Answer 1

Find the initial velocity components ..
Initial vertical vel .. Uv = 25.sin40 .. . Uv = 16.10 m/s
Initial horiz.vel .. Uh = 25.cos 40 .. ... Uh = 19.15 m/s

Find time of flight by applying [ d = Uv.t -

Answer 2

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else)

King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 13 m above the moat. The rock is launched at a speed of 28 m/s and an angle of 32^{\circ} above the horizontal.

How far from the castle wall does the launched rock hit the ground?

Answer

Initial speed of projectile (v) Angle of projection (e)-32 Height of cliff (h)-13m 28m/s (a) Ttaken by projectile to hit point P is got form the vertical motion. 0-13+v sin 32'-49 13+(28.0m/ s)sin32t-49 4.9-14.8377-130 On solving the above equation we get t-3.737879 s The horizontal range is determined for the horizontal motion 0+((28.0m/ s)cos 32) (3.737879) 8.75 m

Answer 3

ducos40 x 3.95 d75. 6 Sm

Answer 4

vo = 15 m/s

vox = vo*cos(40) = 25*cos(40) = 19.15 m/s

voy = vo*sin(40) = 25*sin(40) = 16.07 m/s

let t is the time taken

Now Apply kimetic enquation

-y = voy*t - 0.5*g*t^2

-13 = 16.07*t - 0.5*9.8*t^2

4.9*t^2 - 16.07*t - 13 = 0

by solving the above equation we get,

t = 3.95 s

distance travelled in x direction,

d = vox*t

= 19.15*3.95

= 75.64 m <<<<<<--------------Answer