A ball is thrown with an initial speed v_1 at an angle theta_1 with the horizont

Question

A ball is thrown with an initial speed v_1 at an angle theta_1 with the horizontal. The horizontal range of the ball Is R, and the ball reaches a maximum height R/12. In terms of R and g, find the following. the time interval during which the ball is in motion the ball's speed at the peak of its path the initial vertical component of its velocity its initial speed the angle theta_1 Suppose the ball Is thrown at the same initial speed found in (d) but at the angle appropriate for reaching the greatest height that It can. Find this height. Suppose the ball is thrown at the same initial speed but at the angle for greatest possible range. Find this maximum horizontal range.

Explanation / Answer

a) given that Range is R

maximum height is Hmax = R/12

a) But Range is R = Vi*cos(theta_i)*T = Vi^2*sin^2(theta_i)/g

here T is the time of flight

T = R/(Vi*cos(theta_i))

b) balls's speed at the peak of its path is V = Vi*cos(theta_i) =R/T = R/(2*Vi*sin(theta_i)/g) = R*g/(2*Vi*sin(theta_i))

c)initial vertical component of its velocity is Vi*sin(theta_i) = sqrt(R*g)

d) Vi = sqrt(R*g)/sin(theta_i)

e) we know that

Vi*sin(theta_i) = sqrt(R*g)

sin(theta_i) = sqrt(R*g)/Vi

theta_i = sin^(-1)(sqrt(R*g)/Vi)

f) Hmax = Vi^2/(2*g)

g) xmax = Vi^2/g

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