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Physies 2414-General Physics I Instruc only if your work is ciea parentheses. tions: Answer each question completely, writing

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Answer #1

Given is:-

Height h= 37.8 m

Range  R= 63 m

Angle  heta = 25^circ

Now,

let the initial speed Vo

Then it's components are

v_o_x = v_o cos heta = v_o cos (25)

similarly

Uoy = Vosint-Vosin (25)

Now,

the time taken by the projectile to reach its target is

R = v_o cos (25) imes t

or

t = rac{R}{v_o cos (25)} eq-1

similarly in vertical direction

s = ut + rac{1}{2}at^2

37.8 = (-v_o sin (25))t + rac{1}{2}(9.8)t^2

by plugging the values of t from eq-1 we get

37.8 = (-v_o sin (25))(rac{R}{v_o cos (25)}) + rac{1}{2}(9.8)(rac{R}{v_o cos (25)})^2

which gives us

37.8 =-(63) tan(25)+ (4.9)(63)^2(rac{1}{v_o cos (25)})^2

which gives us

37.8= (-29.38) + rac{19448.1}{v_o^2 imes 0.8214}

which gives us

oxed{v_o = 18.77 m/s} part -b

by plugging this value in eq-1 we get

t= rac{63}{18.77 cos (25)}

which gives us

oxed{t=3.7034 s} part - a

Part -c

Maximum height can be found out by using third equation of motion which is

v^2 = u^2 + 2as

by plugging all the values we get

0^2 = (18.77 sin (25))^2 + 2(-9.8)h_{max}

which gives us

oxed{h_{max} = 3.21 m}

Part - d

The final vertical velocity will be

v_f_y = (-18.77 sin(25)) + (9.8)(3.7034)

which gives us

v_f_y = 28.361 m/s

and

v_f_x = v_o_x = 18.77 cos (25) = 17.01 m/s

thus the final velocity will be

v_f = sqrt{(v_f_x)^2+(v_f_y)^2}

by plugging all the values we get

v_f = sqrt{(28.361)^2+(17.01)^2}

which gives us

oxed{v_f = 33.071 m/s}

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