Question

During a fireworks display, a shell is shot into the air with an initial speed of 91.55

m/s at an angle of theta degrees above the horizontal. The fuse is timed to 8.21

seconds for the shell just as it reaches its highest point above the ground. If

the horizontal displacement of the shell is 149.68 meters when it explodes, what

is the height (in unit of meters) when the shell explodes? You must use g = 9.8

m/s².

Answer 1

Highest point in projectile motion is given by:

H_max = V0y^2/(2*g) = V0^2*(sin )^2/(2*g)

V0 = Initial launching speed = 91.55 m/sec

Range in projectile motion is given by:

R = V0x*t

given that shell is explodes at time 8.21 sec and in this time horizontal distance traveled is 149.68 m, So

V0x = R/t = 149.68/8.21 = 18.23 m/sec

V0*cos = 18.23

= arccos (18.23/91.55) = 78.5 deg = launching angle

Now max height will be:

H_max = V0^2*(sin )^2/(2*g)

H_max = 91.55^2*(sin 78.5 deg)^2/(2*9.81)

H_max = 410.2 m

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