tect is provered at an angle of \( 50^{\circ} \) to the horizontal plane and its initial veloci calculate: he maximum height the object will reach.

To calculate the maximum height achieved by an object projected at a \( 50^{\circ} \) angle to the horizontal, we first need the initial velocity (let's denote it as \( v_0 \)). The vertical component of the initial velocity is given by \( v_{0y} = v_0 \sin(50^{\circ}) \). Using the kinematic equation for vertical motion: \[ h = \frac{v_{0y}^2}{2g} \] where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). Plugging in the expression for \( v_{0y} \): \[ h = \frac{(v_0 \sin(50^{\circ}))^2}{2g} \] This formula will give you the maximum height once you substitute the specific initial velocity value into it! To find the maximum height, just remember the key steps: break down your velocity into components, use the kinematic equations, and enjoy watching your projectile soar higher! If you need specific numbers in, let me know the initial velocity, and I’ll help calculate it!