Answer:
Explanation:
Let the required velocity be V at angle θ with the horizontal . Let time taken in the whole process be t
horizontal component = v cosθ
It will cover .25 m at this velocity
vertical component = v sinθ
It will cover .45 m with this initial velocity and with deceleration g .
v cosθ t = .25
.45 =v sinθ t - 1/2 g t²
putting the values of t from above
.45 = .25 tanθ - g .25² / 2 v² cos²θ
B ) putting the values of θ
.45 = .25 tan63 - 9.8 x .25² / 2 v² cos² 63
.45 = .49 - 1.53 / v²
1.53 / v² = .04
v² = 38.25
v = 6.18 m /s
time taken to cover distance of .25 m
t = .25 / 6.18 cos63 = .089 s
during this period , vertical displacement
= 6.18 sin 63 x .089 - 1/2 x 9.8 x .089²
= .4895 - .04 = .45 m which is consistence with given height .