A projectile is fired with an initial velocity of 88 ft/sec at an angle of 35° to the horizontal. Neglecting air resistance, what is the time of impact? (Round to the nearest tenth of a second.)
Remember: x = (v0 cos θ)t and y = h0 + (v0sinθ)t – 16 t^2
To find the time of impact, we need to determine when the projectile hits the ground. In this case, we can use the equation for the vertical motion of the projectile:
y = h0 + (v0sinθ)t - 16t^2
In this equation:
- y represents the vertical position of the projectile
- h0 is the initial height of the projectile (which we can assume as 0 since it's not mentioned)
- v0 is the initial velocity of the projectile
- θ is the angle at which the projectile is fired
- t is the time
We know the initial velocity (v0 = 88 ft/sec) and the angle (θ = 35°). Since the projectile is fired horizontally from the ground, the initial height (h0) is also assumed to be 0.
Now, let's solve for the time of impact:
Since the projectile hits the ground, the vertical position (y) will also be 0.
So we have:
0 = 0 + (88sin35°)t - 16t^2
Simplifying the equation, we get:
49t - 16t^2 = 0
Since t cannot be 0 (as the projectile needs time to hit the ground), we can divide both sides of the equation by t:
49 - 16t = 0
Solving for t, we get:
16t = 49
t = 49/16
t ≈ 3.06 seconds
Therefore, the time of impact is approximately 3.1 seconds (rounding to the nearest tenth of a second).