Question
Prove that the horizontal range is same when angle of projection is:
- Greater than 45° by certain value.
- Less than 45° by the same value.
then,
$\text{R}_1=\frac{\text{u}^2\sin2(45^\circ+\alpha)}{\text{g}}=\frac{\text{u}^2}{\text{g}}\cos2\alpha\dots(\text{i})$$\therefore\ \text{R}_2=\frac{\text{u}^2\sin2(45^\circ-\alpha)}{\text{g}}=\frac{\text{u}^2}{\text{g}}\cos2\alpha\dots(\text{ii})$
Comparing eq. (i) and (ii), we have R1 = R2Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\text{y}=\frac{\text{a}^3}{(\text{x}-\text{vt})^2+\text{a}^2},$
where a = 5mm and v = 20cm/s. Sketch the shape of the string at t = 0, 1s and 2s.Take x = 0 in the middle of the string.