projectile

Let the ball be thrown at an angle $\theta$ with horizontal.

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$u_x = u\cos\theta \text{ and } u_y = u\sin\theta$

Now, horizontal displacement at any time $t$ is

$x = u_xt = u\cos\theta t$

Now, $x = 5m \text{ and } u = 10m/s$

So,

$5 = (10)\cos\theta t$

$\boxed{t = \frac{1}{2\cos\theta}}\ldots (1)$

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Now, vertical displacement at any time $t$

$y = u_yt - \frac12gt^2$

$y = ((10)\sin\theta)\bigg(\frac{1}{2\cos\theta}\bigg) - \frac12g\bigg(\frac{1}{2\cos\theta}\bigg)^2$

$y = (5\tan\theta) - \frac12(10)\frac{\sec^2\theta}{4}$

$y = (5\tan\theta) - \frac54(1 + \tan^2\theta)$

$y = -\frac54(\tan^2\theta - 4\tan\theta + 1)$

$y = -\frac54((\tan\theta - 2)^2 - 3)$

$y = \frac{15}{4} - \frac54(\tan\theta - 2)^2$

$\text{Maximum value of above function } = \frac{15}{4}$

$\color{#3D99F6}{\boxed{y_{max} = 3.75m}}$