If the initial velocity is 50 meters per second and the launch angle is 15 degrees what is the maximum height? Explain.
Acceleration of the arrow is -3m/s2A = (velocity minus initial velocity) / time
anything shot up with that initial velocity. There isn't anything in specific.
Initial velocity can be measured in the same units as any other velocity. In SI, that would be meters per second, but often km / hour are used, or (in a minority of countries) feet/second or miles/hour.
the answer is 24-9 m/sec. yuor welcome
1 meter/second/second in the same direction of travel
Acceleration of the arrow is -3m/s2A = (velocity minus initial velocity) / time
anything shot up with that initial velocity. There isn't anything in specific.
You kicked the rock with an initial velocity of 3.4 m/s.
Initial velocity can be measured in the same units as any other velocity. In SI, that would be meters per second, but often km / hour are used, or (in a minority of countries) feet/second or miles/hour.
Acceleration occurs when velocity changes over time. The formula for it is as follows: a = (Vf - Vi) / t a: acceleration (meters/seconds2) Vf: Final velocity (meters/seconds) Vi: Initial Velocity (meters/seconds) t: Time (seconds)
the answer is 24-9 m/sec. yuor welcome
In the usual simple treatment of projectile motion, the horizontal component of the projectile's velocity is assumed to be constant, and is equal to the magnitude of the initial (launch) velocity multiplied by the cosine of the elevation angle at the time of launch.
It will depend upon the initial velocity of the body. If 'u' be the initial velocity of the body, then the final velocity will be: v = u + at (v = final velocity, a = acceleration, t = time) i.e., v=u+10*7 = (u + 70) m/sec. If u=0 (i.e the initial velocity be zero) then final velocity, v=70 m/sec.
a=change over velocity/time 60-initial velocity 45-final velocity 45-60= 15m/s 15/5= 3- acceleration
You cannot. You need to know either the initial speed or angle of projection (A).
0.82 metres.
0.82 metres.