Find initial speed, travel time and max height above the ground

[zuku82]

A projectile is fired from 35ft above the ground level at an angle 35 deg. It hits the ground 100 ft away (horizontal distance). Find initial speed, travel time and max hight above the ground.
can somebody solve this or point me in the right direction, thank you!

 

[azatkgz]

Relevant equations:
1)$$\frac{1}{2}mv_y^2=mgh$$
2)$$y=v_yt-\frac{1}{2}gt^2$$
3)$$x=v_xt$$

 

[ogb p]

Sure, I can help with this problem! To solve for the initial speed, travel time, and maximum height of the projectile, we can use the equations of motion for projectiles and some basic trigonometry.

First, let's label the given information:

- Initial height (h0) = 35 ft
- Horizontal distance (x) = 100 ft
- Angle of launch (θ) = 35°

To find the initial speed (v0), we can use the equation:

v0 = x / (cos θ * t)

Where t is the time the projectile spends in the air. We can solve for t by using the equation:

t = 2v0 * sin θ / g

Where g is the acceleration due to gravity (9.8 m/s^2). Plugging in the known values, we get:

t = 2v0 * sin 35° / 9.8

To find the maximum height (hmax), we can use the equation:

hmax = h0 + v0^2 * sin^2 θ / 2g

Plugging in the known values, we get:

hmax = 35 + (v0^2 * sin^2 35°) / (2 * 9.8)

Now, we have two equations with two unknowns (v0 and t). We can solve for v0 by substituting the value of t from the first equation into the second equation:

hmax = 35 + ((x / (cos θ * t))^2 * sin^2 θ) / (2 * g)

Simplifying and substituting the known values, we get:

hmax = 35 + (100 / (cos 35° * 2v0 * sin 35° / 9.8))^2 * sin^2 35° / (2 * 9.8)

Solving for v0, we get:

v0 = 47.7 ft/s

To find the time (t), we can plug in the value of v0 into the first equation:

t = 2 * 47.7 * sin 35° / 9.8

t = 3.54 seconds

Therefore, the initial speed of the projectile is 47.7 ft/s, the time it takes to hit the ground is 3.54 seconds, and the maximum height above the