Answer to Question #304284 in Mechanics | Relativity for KING

Question #304284

A batter hits a baseball so that it leaves the bat with an initial speed 37.0 an initial angle 53.1o, at a location. Find

a) the position of the ball at t = 2.00 s.

b) the magnitude and direction of its velocity,

when t = 2.00 s.

c) the time when the ball reaches the highest point of its flight. d) its maximum height, h.

e) its horizontal range, 𝑅.


1
Expert's answer
2022-03-06T18:15:36-0500

u=37m/s

θ=53.1°\theta=53.1°

Now

s=u(cosθ)t+12at2s=u(cos\theta) t+\frac{1}{2}at^2

Put value


s=37×cos53.1×2+0.5×9.8×22s=37\times cos53.1\times2+0.5\times9.8\times2^2

s=64.03sec

v2=u2+2asv^2=u^2+2as

v2=(ucosθ)2+2×a×sv^2=(ucos\theta)^2+2\times a\times s


v2=37×cos53.1+2×9.8×64.03=1748.52v^2=37\times cos53.1+2\times9.8\times64.03=1748.52

v=41.82m/secv=41.82m/sec

t=usinθgt=\frac{usin\theta}{g}

t=37×sin53.19.8=3.02sect=\frac{37\times sin53.1}{9.8}=3.02sec

Max height

H=u2sin2θ2gH=\frac{u^2sin^2\theta}{2g}

H=372sin2(53.1)2×9.8=44.66mH=\frac{37^2sin^2(53.1)}{2\times9.8}=44.66m

Range

R=u2sin2θgR=\frac{u^2sin2\theta}{g}

R=372sin106.29.8=134.15mR=\frac{37^2sin106.2}{9.8}=134.15m


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