Question

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

Answer 1

Answer:

Step-by-step explanation:

We will work in the y-dimension only here. What we need to remember is that acceleration in this dimension is -9.8 m/s/s and that when the projectile reaches its max height, it is here that the final velocity = 0. Another thing we have to remember is that an object reaches its max height exactly halfway through its travels. Putting all of that together, we will solve for t using the following equation.

$v=v_0+at$

BUT we do not have the upwards velocity of the projectile, we only have the "blanket" velocity. Initial velocity is different in both the x and y dimension. We have formulas to find the initial velocity having been given the "blanket" (or generic) velocity and the angle of inclination. Since we are only working in the y dimension, the formula is

$v_{0y}=V_0sin\theta$ so solving for this initial velocity specific to the y dimension:

$v_{0y}=35sin(35)$ so

$v_{0y}=$ 2.0 × 10¹ m/s

NOW we can fill in our equation from above:

0 = 2.0 × 10¹ + (-9.8)t and

-2.0 × 10¹ = -9.8t so

t = 2.0 seconds

This is how long it takes for the projectile to reach its max height. It will then fall back down to the ground for a total time of 4.0 seconds.