Question

An angry bird is thrown from a slingshot in 2.50 m height with the starting speed of 15.9 m/s. The bird lands on a naughty pig in 3.60 m height and 14 m horizontal distance from the slingshot.

a) calculate the shooting angle i.e. the direction of the
starting speed of the bird. You may use iteration to calculate, or
the equtaion 1/cos^{2}*?* =
1+tan^{2}*?.* Is one angle that is possible or mor?
discuss.

b) What is the speed of the bird when it lands on the pig? (magitude and direction)? If you got two angles from part a) then use the bigger angle.

Help please :)

Answer #1

x direction

x = v0x t

14 = 15.9*cos theta* t

t = 14/(15.9*cos(theta))

y direction

y = y0 + v0 t + 1/2 a t^2

3.6 = 2.5 + 15.9*sin(theta)*t - 0.5*9.81*t^2

plug in t

3.6 = 2.5 + 15.9*sin(theta)*14/(15.9*cos(theta)) -
0.5*9.81*(14/(15.9*cos(theta)))^2

use that 1/cos^2 = 1 + tan^2

3.6 = 2.5 + 15.9*tan(theta)*14/(15.9) - 0.5*9.81*(1 + tan(theta)^2)(14/(15.9)^2

now solve for tan(theta)

tan(theta) = 0.09816 or 51.443

theta = arctan(0.09816)= 5.6 degrees

or theta = arctan(51.443)= 88.89 degrees

b)

vx doesnt change, vx = 15.9*cos(88.89 degrees)=0.308 m/s

for vy

vy^2 = v0^2 + 2 a dy

vy = sqrt( (15.9*sin(88.89 degrees))^2 -2*9.81*(3.6-2.5))=15.2 m/s

mag = sqrt(vx^2 + vy^2) = sqrt(15.2^2 + 0.308^2)= 15.2 m/s

angle = arctan(y/x) = arctan(15.2/0.308)= 88.84 degrees

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