Agray kangaroo can bound across level ground with each jump carrying it 9.6 m from the takeoff point. typically the kangaroo leaves the ground at a 28º angle. if this is so: part a) what is its takeoff speed? express your answer to two significant figures and include the appropriate units. part b) what is its maximum height above the ground? express your answer to two significant figures and include the appropriate units

(A) 11 m/s

(B) 1.3 m

Explanation:

Horizontal range, R = 9.6 m

Angle of projection, theta = 28 degree

(A)

Use the formula of horizontal range

R = u^2 Sin 2 theta / g

u^2 = R g / Sin 2 theta

u^2 = 9.6 × 9.8 / Sin ( 2 × 28)

u = 10.65 m/s

u = 11 m/s

(B)

Use the formula for maximum height

H = u^2 Sin ^2 theta / 2g

H =

10.65 × 10.65 × Sin^2 (28) / ( 2 × 9.8)

H = 1.275 m

H = 1 .3 m

dang that balls went far

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards