Respuesta :
The speed of his take off is 43.2 m/s if he left the take-off ramp at 120° to the horizontal and that the take-off and landing heights are the same. neglecting air drag.
Explanation
We have given that
R = 77.0m
[tex]\theta_0[/tex] = 12.0°
The horizontal range of such projectile R is given from the relation :
[tex]R = \frac{v^2_0}{g} sin 2 \theta_0[/tex]
Here, [tex]v_0[/tex] is velocity of take-off & [tex]\theta_0[/tex] is the angle of take-off
With this in mind,
[tex]v_0[/tex] = [tex]\sqrt{\frac{gR}{sin2\theta_0}[/tex]
= [tex]\sqrt{\frac{(9.8 m/s^2)(77m)}{sin24}[/tex]
= 43.2 m/s
What is velocity?
Velocity is the directional speed of a moving item as an indication of its rate of change in position as seen from a specific frame of reference and measured by a specific time standard (e.g., 60 km/h northbound). The idea of velocity is crucial in kinematics, the part of classical mechanics that explains the motion of bodies.
A physical vector quantity called velocity requires both magnitude and direction to be defined. Speed is a coherent derived unit that expresses the scalar absolute value (magnitude) of velocity. Its quantity is expressed in SI (metric system) units of metres per second (m/s or ms1). For example, "5 meters per second" is a scalar while "5 meters per second east" is a vector.
Learn more about Velocity
https://brainly.com/question/80295?source=archive
#SPJ4
