Question

A gray 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

Answer 1

Answer:

(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

Answer 2

(a)The take-off speed is the speed at the start of takeoff. The take-off speed of the kangaroo will be 11 m/sec.

(b)The height achieved during takeoff is the maximum height. The maximum height above the ground will be 1.3 meters.

what is the maximum height achieved in projectile motion?

It is the height achieved by the body when a body is thrown at the same angle and the body is attaining the projectile motion.The maximum height of motion is given by

$H = \frac{u^{2}sin^2\theta }{2g}$

What is a range of projectile?

The horizontal distance is covered by the body when the body is thrown at some angle is known as the range of the projectile. It is given by the formula

$R = \frac{u^{2}sin2\theta}{g}$

(a)Take-of velocity =?

given

Horizontal range = 9.6m.

$\theta = 28^0$

$g = 9.81 \frac{m}{sec^{2} }$

$R = \frac{u^{2}sin2\theta}{g}$

$u = \sqrt{\frac{Rg}{sin2\theta} }$

$u = \sqrt{\frac{9.6\times9.81}{sin56^0} }$

$u = 11 m /sec$

Hence the take-off speed of the kangaroo will be 11 m/sec.

(b) Maximum height =?

given,

$u = 11 m /sec$

$H = \frac{u^{2}sin^2\theta }{2g}$

$H = \frac{(11)^{2}sin^2 58^0 }{2\times 9.81}$

$\rm { H = 1.3 meter }$

Hence the maximum height above the ground will be 1.3 meters.

To learn more about the range of projectile refer to the link ;

https://brainly.com/question/139913