a = -g = -9.81 m/s^2
All that matters here is the vertical part of the problem.
Vi =initial speed up = 12 sin 28 = 5.63 m/s
v = Vi - 9.81 t
at top v = 0
so
t = 5.63 /9.81 = 0.574 second
h=Hi + Vi t - (9.81/2) t^2 = 0 + 5.63*.574 - 4.9*.574^2
= 3.23 - 1.61 = 1.61 meters hi
If you wanted to know where it lands do constant horizontal velocity
u = 12 cos 28 forever
time =2 t = 2 * .574
range = 2 * .574 * 12 cos 28
= Vi cos
A long jumper leaves the ground with an initial velocity of 12 m/s at an angleof 28°
above the horizontal. Determine the peak height of the long jumper?.
1 answer