Question

solve completely. explain your answer please

Answer 1

a)

Let +x be towards leftward and +y be towards upward

X and Y components of Final Horizontal velocity

V_fx=VCos60

V_fy=-VSin60

Here final horizontal velocity is equal to initial horizontal velocity i.e

V_fx=V_ox

Horizontal distance traveled

d=V_fxt

25 =(VCos60)(1.5)

V=33.33 m/s

Therefore

V_fx=V_ox=33.33Cos60 =16.67 m/s

V_fy=-33.33Sin60=-28.87 m/s

From

V_fy=V_oy-gt

-28.87 =V_oy -9.81*1.5

V_oy=-14.1525 m/s

Since

$V_{ox}=V_{o}Cos\phi =16.67$----------------1

$V_{oy}=V_{o}SIn\phi =-14.1525$------------2

2/1

$tan\phi =\frac{-14.1525}{16.67}=>\phi =tan^{-1}(0.849)=-40.34^{o}$

Therefore

From 1

V_o=16.67/Cos(-40.34) =21.87 m/s

From

Y=Y_o+V_oyt-(1/2)gt²

when the ball hits ground

0=h+(-14.1525)(1.5)-(1/2)(9.81)(1.5)²

h=32.27 m

b)

V_o=21.87 m/s

c)

$\phi =-40.34^{o}$

d)

Since angle is negative ,ball is thrown below horizontal