Question

A small, spring-loaded cannon launches a golf ball from level ground with an initial speed v_i at an angle θ_i with the horizontal. The golf ball lands a horizontal distance R from its launch point. The highest point the golf ball reaches during its flight is a distance R/12 above the ground. In terms of R and g, find the following. (You may ignore air resistance.)

(a)

the time interval during which the golf ball is in motion

​​​​​t =

(b)

the golf ball's speed at the peak of its path

v =

(c)

the initial vertical component of its velocity

v_yi =

ONLY 3 PARTS FOR CREDIT. REST FOR PRACTICE.

(d)

its initial speed

(e)

the angle θ_i (Assume θ_i is in radians. Do not include units in your answer.)

(f)

Suppose the golf ball is launched at the same initial speed found in (d) but at the angle appropriate for reaching the greatest height that it can. Find this height.

(g)

Suppose the golf ball is launched at the same initial speed but at the angle for greatest possible range. Find this maximum horizontal range.

Answer 1

Given given, initial relouty of her projection = n Angle of projection, Op. Range =R, maximum height, Hmax e (a) Time Entequal in which golf ball is in moias is time of flight e T = 24 sm o =) T= av sine seconds g

we know, range, R = u Sin 20; - Sus 20; 9 g = v = Rg Rg = Ve. Sin 20. 2 Sino costi T= 2u sunde = a (ne sander) but R Tano: = 4 tmax =) Tand= 4 (2) =) T = 2 Sind Rg g 12. Suso. Cosa R =) Tano = 1 / 2 = 4 Simo; l&g) 9-12 Sim O; Coso.) V T= [2R tanki = 28 (+3) = 139 (b) At peak of its path, its speed is horizontal Component of initial speed which remains constart! through out the motion. v=vo coroo = Cosle 14 Carlo Rg 2 Sim O Cosa = v= Rg cato, - eg (3) = 3Rg kn 13 13Rg V 2 2

(c) gritual vertical component of relocity is Vyi = Ve Sinoo & Sunde / RgT V asiso, coso; voi = Rg tandi = Rg (3) Vaji = v Rog It's initial Speed Vo is given by (d) =) 11= V = 1 (329) + () 100g 6 v = 5.99 (e) for angle of projection, Tan Oo = vai 5) Tands on 1,6)=5707 Tail =)

(f) for maximum height, 0290 Then Maximum height reached is. Hmax = & sing 29 5Rg Hmax = 58 (9) for Range to be maximum, og = 45° - Rmax = 1 Sin 245) - Rmax - Se