Update app.py

app.py

@@ -1,26 +1,29 @@
 import gradio as gr
 
-gr.Markdown("""
-The range of a projectile is the horizontal distance it travels during its motion. The horizontal displacement of a projectile at any time can be calculated using the following kinematic equation:
-
-$$x = x_0 + v_0 \cos \theta \cdot t$$
-
-where $x$ is the horizontal displacement, $x_0$ is the initial horizontal position, $v_0$ is the initial velocity, $\theta$ is the angle at which the projectile is launched, and $t$ is the time.
-
-To find the range of the projectile, we need to find the time at which it hits the ground (i.e., when its vertical displacement becomes zero). The vertical displacement of a projectile at any time can be calculated using the following kinematic equation:
-
-$$y = y_0 + v_0 \sin \theta \cdot t - \frac{1}{2}gt^2$$
-
-where $y$ is the vertical displacement, $y_0$ is the initial vertical position, $v_0$ is the initial velocity, $\theta$ is the angle at which the projectile is launched, $t$ is the time, $g$ is the acceleration due to gravity, and $t$ is the time.
-
-To find the time at which the projectile hits the ground, we can set the vertical displacement to zero and solve for $t$. This gives us the following equation:
-
-$$0 = y_0 + v_0 \sin \theta \cdot t - \frac{1}{2}gt^2$$
-
-Solving for $t$, we get:
-
-$$t = \frac{v_0 \sin \theta \pm \sqrt{v_0^2 \sin^2 \theta + 2gy_0}}{g}$$
-
-Since the projectile will hit the ground at a later time, we need to take the positive value of $t$. Substituting this value into the equation for horizontal displacement, we get the following equation for the range of the projectile:
-
-""").launch()
+with gr.Blocks() as demo:
+    gr.Markdown("""
+    The range of a projectile is the horizontal distance it travels during its motion. The horizontal displacement of a projectile at any time can be calculated using the following kinematic equation:
+    
+    $$x = x_0 + v_0 \cos \theta \cdot t$$
+    
+    where $x$ is the horizontal displacement, $x_0$ is the initial horizontal position, $v_0$ is the initial velocity, $\theta$ is the angle at which the projectile is launched, and $t$ is the time.
+    
+    To find the range of the projectile, we need to find the time at which it hits the ground (i.e., when its vertical displacement becomes zero). The vertical displacement of a projectile at any time can be calculated using the following kinematic equation:
+    
+    $$y = y_0 + v_0 \sin \theta \cdot t - \frac{1}{2}gt^2$$
+    
+    where $y$ is the vertical displacement, $y_0$ is the initial vertical position, $v_0$ is the initial velocity, $\theta$ is the angle at which the projectile is launched, $t$ is the time, $g$ is the acceleration due to gravity, and $t$ is the time.
+    
+    To find the time at which the projectile hits the ground, we can set the vertical displacement to zero and solve for $t$. This gives us the following equation:
+    
+    $$0 = y_0 + v_0 \sin \theta \cdot t - \frac{1}{2}gt^2$$
+    
+    Solving for $t$, we get:
+    
+    $$t = \frac{v_0 \sin \theta \pm \sqrt{v_0^2 \sin^2 \theta + 2gy_0}}{g}$$
+    
+    Since the projectile will hit the ground at a later time, we need to take the positive value of $t$. Substituting this value into the equation for horizontal displacement, we get the following equation for the range of the projectile:
+    
+    """)
+    
+demo.launch()