Question

A projectile is fired with initial speedv -100 feet per second from a height of h 0 feet atan angle of θ-7/6 above the horizontal. Assuming that the only force acting on the object is gravity, find the maximum altitude, horizontal range and speed at impact.

Answer 1

Answer:

Maximum altitude: 497.96 ft

Horizontal range: 1007.37 ft

Speed at impact:  165.21 ft/s

Step-by-step explanation:

angle(α) = atan (7/6) = 49.4°

Maximum altitude is given by the formula:

$h=\frac{V_0^2sin^2\alpha }{2g}$

$h=\frac{100^2 sin^2(49.4)}{2*9.81} =\frac{9770}{19.62}=497.96 ft/s$

Horizontal range is given by the formula:

$X=\frac{V_0^2sin(2\alpha)}{g}$

$X=\frac{100^2sin(2*49.4)}{*9.81}=1007.37 ft$

Speed at impact is given by the formula:

$V_f=\sqrt{V_x^2 + Vy^2}$

where:

$V_x= V_0cos(\alpha )= 100cos(49.4)=65.07 ft/s$

$V_y=V_0sin(\alpha ) + gt=100sin(49.4)+9.81(t)$

$t=\frac{V_0sin(\alpha) }{g}=\frac{100sin(49.4)}{9.81}=7.74s$

So;

$V_y= 100sin(49.4)+(9.81)(7.74)= 151.86 ft/s$

$Vf=\sqrt{V_x^2 + V_y^2} =\sqrt{65.07^2+151.86^2}=165.21 ft/s$