Respuesta :
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
(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
[tex]H = \frac{u^{2}sin^2\theta }{2g}[/tex]
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
[tex]R = \frac{u^{2}sin2\theta}{g}[/tex]
(a)Take-of velocity =?
given
Horizontal range = 9.6m.
[tex]\theta = 28^0[/tex]
[tex]g = 9.81 \frac{m}{sec^{2} }[/tex]
[tex]R = \frac{u^{2}sin2\theta}{g}[/tex]
[tex]u = \sqrt{\frac{Rg}{sin2\theta} }[/tex]
[tex]u = \sqrt{\frac{9.6\times9.81}{sin56^0} }[/tex]
[tex]u = 11 m /sec[/tex]
Hence the take-off speed of the kangaroo will be 11 m/sec.
(b) Maximum height =?
given,
[tex]u = 11 m /sec[/tex]
[tex]H = \frac{u^{2}sin^2\theta }{2g}[/tex]
[tex]H = \frac{(11)^{2}sin^2 58^0 }{2\times 9.81}[/tex]
[tex]\rm { H = 1.3 meter }[/tex]
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
