What is the max hieght and range for a ball launched at 15 degrees at 20 m/s. 20 degrees at 20m/s. 30 degrees at 20 m/s. 45 degrees at 20 m/s. 50 degreesat 20 m/s. 75 degrees at 20 m/s. 90 degrees at 20m/s.30 degrees at 30 m/s. 50 degreesat 50 m/s.
Use the following formula to learn how to do it. You have to use a calculator:h = v(sin theta)t + 1/2 (-9.8) t^2, where h=height, t=time in seconds. Good luck.
+130536 Here's the first one....alll the rest are just the same idea.......
The vertical position function is : h(t) = vsin(theta) t - (4.9)t^2 where v is the original velocity, theta is the launch angle, and h is the height at any time
So, taking the derivative of this and setting it to 0, we have
h'(t) = 20sin(15) - (9.8)t = 0 and solving this for t, we have.....t = 0.528202 seconds......
And this is the time to reach the max ht
And that height = 20sin(15)(0.528202) - (4.9)(0.528202)^2 = 1.36709m
And the horizontal displacement (range) = v * t , where v is the intital velocity and t is the total time..... 20m/s * (2*0.528202)s = 21.128m
