A projectile is fired from ground level with an initial velocity of 35 m/s at an angle of 35° with the horizontal. How long will it take for the projectile to reach the ground? (Show work

Answer 1

Hi Chelsea,

The formula for the height of the projectile is

y = v₀ * sin(θ) * t - (1/2) * g * t²

Where v₀ = initial velocity, θ = initial angle, t = time, and g = 9.8 (force of gravity). Plugging these in we get:

y = 35 * sin(35) * t - 4.9 * t²

We want to know when it hits the ground, meaning when y = 0.

0 = 35 * sin(35) * t - 4.9 * t²

Now we can factor out a t on the right side:

0 = t * (35 * sin(35) - 4.9 * t)

One solution for this is t = 0 which is when the projectile is launched. We want the other, when

0 = 35 * sin(35) - 4.9 * t

4.9 * t = 35 * sin(35)

t = 35 * sin(35) / 4.9 = 4.1 seconds.

Please let me know if I can be more helpful, thanks!

Joe