Question

What is the solution of this question?

Answer 1

V = √[Rmax*g] = √[240*9.8] = 48.5 m/s
for Rmax = 240m ; θ = 45 deg; let vel = u 
v=u-gt 
at max height, 
0 = u sin 45 - 9.81* t1 
& 
t1 = 120 / u cos 45 
u = 120 / t1 cos 45 
[120 / t1 cos 45 ] * sin 45 = 9.81 *t1 
t1^2 = [120/ 9.81] * tan 45 
t1 = 3.49 s 
tmax = 2 * t1 = 6.98 s 

for 0.4 tmax = 2.792 s, at angle A; same range 240m 
at max height, t = 2.792 / 2 = 1.396 s 
0 = u sin A - 9.81* 1.396 
u sin A = 13.69 

1.396 = 120 / u cos A 
u cos A = 85.96 

tan A = 0.1592 
A =9.045 deg 
u = 13.69 / sin 9.045 
u = 87.08 m/s

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