Question

1. an object thrown horizontally from a height h with initial speed ( v_0 ) strikes the ground with a velocity of 50 m/s, making an angle of ( 53^{circ} ) with the horizontal axis, as shown in the figure. what is h, in metres? (take ( sin 53^{circ}=0.8 ); ( cos 53^{circ}=0.6 ); ( g = 10 mathrm{~m} / mathrm{s}^2 )) a) 45 b) 60 c) 80 d) 100 e) 125

Explanation:

Step1: Find the vertical component of the final velocity

The final velocity \(v = 50\ m/s\) and the angle \(\theta=53^{\circ}\) with the horizontal.
The vertical component of the final velocity \(v_y\) is given by \(v_y = v\sin\theta\).
Substituting the values: \(v_y=50\times0.8 = 40\ m/s\).

Step2: Use the kinematic equation for vertical motion

For vertical motion, the kinematic equation \(v_y^{2}=u_y^{2}+2gh\).
Since the object is thrown horizontally, the initial vertical velocity \(u_y = 0\ m/s\).
The equation simplifies to \(v_y^{2}=2gh\).
We need to solve for \(h\), so \(h=\frac{v_y^{2}}{2g}\).
Substituting \(v_y = 40\ m/s\) and \(g = 10\ m/s^{2}\):
\(h=\frac{40^{2}}{2\times10}=\frac{1600}{20}=80\ m\).

Answer:

C. 80