To find the horizontal range of a ball projected upwards with a velocity of 25 m/s at an angle of $35^\circ$ with the horizontal, you can use the formula for the horizontal range $R$ of a projectile:

$R = \frac{v^2 \sin 2\theta}{g}$

Where:

• $v = 25 \, \text{m/s}$ is the initial velocity,
• $\theta = 35^\circ$ is the angle of projection,
• $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.

Substitute the given values into the formula:

$R = \frac{(25)^2 \sin(2 \times 35^\circ)}{9.8}$

Calculate $\sin(70^\circ)$:

$\sin(70^\circ) \approx 0.9397$

Now, substitute and solve:

$R = \frac{625 \times 0.9397}{9.8}$

$R \approx \frac{587.3125}{9.8} \approx 59.93 \, \text{m}$

The horizontal range of the ball is approximately $59.93 \, \text{m}$.