A quarterback is about to throw a pass to his wide receiver running a route. The
ID: 3886110 • Letter: A
Question
A quarterback is about to throw a pass to his wide receiver running a route. The quarterback is 6 ft tall and is supposed to hit the wide receiver straight down the field 25 yards away. The equation that describes the motion of the football is the familiar equation of projectile motion from physics, y = x tan(0) - 1/2 x^2 g/v^2_0 1/cos^2(theta) + h where x & y are the horizontal and vertical distances, respectively, V_0 is the initial velocity of the football as it leaves the 60 ft quarterback's hand and theta is the angle the football makes with the ground as it leaves the quarterback's hand. For v_0 = 50 ft/s, x = 20 yds, h = 6ft & y = 7ft, find the angle at which the quarterback must launch the ball. Use bisect.p Print the answer to the screen in degrees, accurate to 2 sig figs, using fprintf.Explanation / Answer
Please refer below code
close all
clear all
clc
v0 = 50; %ft/s
x = 20 * 3.0; %yard to ft
h = 6; %ft
y = 7; %ft
g = 9.8/0.3048; %m/s^2 to ft/s^2
%wrote equation with F = 0 i.e. 0 on RHS and other terms on LHS
f = @(theta)(x * tan(theta) - ((x^2 * g)/(2 * (v0 * cos(theta))^2)) + h - y);
theta = fzero(f,1);
fprintf('theta = %.2f ', theta);
Answer:
theta = 1.13