Answer to Question #304284 in Mechanics | Relativity for KING

Question

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.

Answer 1

u=37m/s

$\theta=53.1°$

Now

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

Put value

$s=37\times cos53.1\times2+0.5\times9.8\times2^2$

s=64.03sec

$v^2=u^2+2as$

$v^2=(ucos\theta)^2+2\times a\times s$

$v^2=37\times cos53.1+2\times9.8\times64.03=1748.52$

$v=41.82m/sec$

$t=\frac{usin\theta}{g}$

$t=\frac{37\times sin53.1}{9.8}=3.02sec$

Max height

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

$H=\frac{37^2sin^2(53.1)}{2\times9.8}=44.66m$

Range

$R=\frac{u^2sin2\theta}{g}$

$R=\frac{37^2sin106.2}{9.8}=134.15m$