2 marks A boy at O throws a ball in the air with a speed vo

. (2 marks) A boy at O throws a ball in the air with a speed vo at an angle . If he then throws another ball at the same speed vo at an angle 6,

along horizontal

initial velocity vox = vo*costheta

acceleration ax = 0

displacement = vox*t + (1/2)*ax*t^2

for ball 1

x = vo*cos45*t1

for ball 2

x = vo*cos30*t2

vo*cos45*t1 = vo*cos30*t2

t2 = (cos45/cos30)*t1 .............(1)

along vertical

initial velocity voy = vo*sintheta

acceleration ay = -g

displacement dy = y

from equation of motion

dy = voy*t + (1/2)*ay*t^2

for ball1

y = vo*sin45*t1 - (1/2)*g*t1^2

for ball 2

y = vo*sin30*t2 - (1/2)*g*t2^2

vo*sin45*t1 - (1/2)*g*t1^2 = vo*sin30*t2 - (1/2)*g*t2^2

(t1*sin45 - t2*sin30)*vo = (1/2)*g*(t1^2-t2^2)

(t1*sin45 - t2*sin30)*10 = (1/2)*9.8*(t1^2-t2^2)

from 1

(t1*sin45 - (cos45/cos30)*sin30*t1) = (1/20*9.8*(t1^2 - ((cos45/cos30*t1)^2)

t1 = 1.83 s

t2 = 1.49 s

time between the throws = t1 - t2 = 0.34 s

DONE please check the answer. any doubts post in comment box