Answer
$v_A=19.4$ m/s
$t_{AB}=4.54$ s
Work Step by Step
$s=v_0t$
$100(\frac{4}{5})=v_a\cos 25t_{AB}$
$s=s_0+v_0t+\frac{1}{2}a_ct^2$
$-4-100(\frac{3}{5})=0+v_A\sin 25t_{AB}+\frac{1}{2}(-9.81)t^2_{AB}$
Solving
$v_A=19.4$ m/s
$t_{AB}=4.54$ s
Engineering Mechanics: Statics & Dynamics (14th Edition)
by Hibbeler, Russell C.
Published by
Pearson
ISBN 10:
0133915425
ISBN 13:
978-0-13391-542-6
Chapter 12 - Kinematics of a Particle - Section 12.6 - Motion of a Projectile - Problems - Page 53: 96
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