Video Transcript
A projectile is fired at an angle
of 66 degrees above the horizontal. The time between the projectile
leaving the ground and returning to the ground at the same height that it was
launched from is 2.9 seconds. What was the projectile’s Initial
speed?
Okay, let’s say that this is the
path that our projectile follows. The initial launch angle of this
projectile, we’ll call it 𝜃, is 66 degrees. And the time it takes for our
projectile to travel this full path, we’ll call this 𝑡 sub 𝑓, is 2.9 seconds. Knowing this, we want to solve for
the initial speed of the projectile. We’ll call it 𝑣 sub 𝑖. Since we are working with a
projectile, a body moving under the influence only of the force of gravity, we can
recall that the total time it takes for a trajectory that starts and ends at the
same height is equal to two times the projectile’s initial speed multiplied by the
sin of its launch angle all divided by 𝑔.
In our situation, it’s not 𝑡 sub
𝑓 we want to solve for but 𝑣 sub 𝑖. To begin doing that, we can
multiply both sides of this equation by 𝑔 over two times the sin of 𝜃. Over on the right-hand side, the
factors of two, 𝑔, and sin 𝜃 all cancel out, leaving us with 𝑣 sub 𝑖. And we see then that 𝑣 sub 𝑖 is
equal to 𝑔 times 𝑡 sub 𝑓 over two times the sin of 𝜃.
Looking at what’s given to us in
our problem statement, we know 𝜃 as well as 𝑡 sub 𝑓, and we can recall further
that 𝑔 equals 9.8 meters per second squared. If we then plug in these values to
our equation for 𝑣 sub 𝑖, to two significant figures, we get a result of 16 meters
per second. This is the initial speed our
projectile would need to have at this given launch angle so that it’s in the air for
2.9 seconds.
