Barky the physics dog is able to leap at a speed of 200 m/s at an angle of 35 degrees,

Question

a) What is the maximum horizontal distance Barky can leap?
b) If Barky wanted to double his horizontal distance without changing his velocity or angle, from what height must he jump from?

Answer 1

initial vertical speed: 200sin35
time in air: hf=hi+vi't-4.9t^2
0=0+200sin35deg*t-4.9t^2
t= 200sin35/4.9
Horizontal distance=200cos35*t

well, if horizontal distance is doubled, t must be doubled, so
0=hi+200sin34*2t-4.9(2t)^2 solve for hi

Answer 2

u = 200 cos 35, forever, for parts a and b both
(that means he will have to stay in the air twice as long for b)

a) Vi = 200 sin 35
how long going up?
v = Vi - g t
v = 0 at the top
so time up Vi/g = 200 sin 35 / 9.81
total time = twice that = 400 sin 35 / 9.81
so horizontal distance = 200 cos 35 * 400 sin 35 / 9.81

b)
new time in air = twice old time = t = 800 sin 35/9.81
h = Hi + Vi t - 4.9 t^2
0 = Hi + 200 sin 35* 800 sin 35/9.81 - 4.9 (800 sin 35/9.81)^2
solve for Hi, initial height