Answer:
![s=203.149\ m](https://img.qammunity.org/2021/formulas/physics/college/t80t3r00pu8spregglwzzibtqd4py4fzhj.png)
Step-by-step explanation:
Given:
- initial velocity of projectile,
![u=41\ m.s^(-1)](https://img.qammunity.org/2021/formulas/physics/college/d2ulywyy6fe3q2pbyupee81cun4lvmlccv.png)
- angle of projection above horizontal,
![\theta=38^(\circ)](https://img.qammunity.org/2021/formulas/physics/college/uv33rip2ir13jd0q7j7zjk9injbju82j8h.png)
height of the initial projection point above the ground,
![y=35\ m](https://img.qammunity.org/2021/formulas/physics/college/dvbsfqd1xd4rhfd7mprgini3gpaxhx8gor.png)
Vertical component of the velocity:
![u_y=u.\sin\theta](https://img.qammunity.org/2021/formulas/physics/college/pwor2irpynofgjw92apocvl4go7lesvk4p.png)
![u_y=41* \sin38](https://img.qammunity.org/2021/formulas/physics/college/ohbgcn5l8uuv7wp2o0k6njerupiixihvfe.png)
![u_y=25.242\m.s^(-1)](https://img.qammunity.org/2021/formulas/physics/college/mj6s4boc3hoc1h497n4qjwj2gmw0xrt871.png)
The time taken in course of going up:
(at top the final velocity will be zero)
![v_y=u_y-g.t](https://img.qammunity.org/2021/formulas/physics/college/4isu6eyosn86hl1kybsxr3e9xscjhptlw4.png)
![0=25.242-9.8* t](https://img.qammunity.org/2021/formulas/physics/college/kgxzgqwahb9ginuri37voc7jn6zs5knqcr.png)
![t=2.576\ s](https://img.qammunity.org/2021/formulas/physics/college/nrryj5y3h3v81firv8s3yaqankeo3kbic2.png)
In course of going up the maximum height reached form the initial point:
(at top height the final velocity is zero. )
using eq. of motion,
![v_y^2=u_y^2-2* g.h](https://img.qammunity.org/2021/formulas/physics/college/aq9dq99vy17n3oagjim9eemal9ii0vbrme.png)
where:
final vertical velocity while going up.=0
maximum height
![0^2=25.242^2-2* 9.8* h](https://img.qammunity.org/2021/formulas/physics/college/4cyaemgwlvvbtx0mo977r796unhjz515kl.png)
![h=32.5081\ m](https://img.qammunity.org/2021/formulas/physics/college/t53kx6c9kw0cgi9m1o17lykf3mft1hy6et.png)
Now the total height to be descended:
![h'=h+y](https://img.qammunity.org/2021/formulas/physics/college/cr46pzmwlql7e60tjjfr3zhmzlu9b8q0gv.png)
![h'=32.5081+35](https://img.qammunity.org/2021/formulas/physics/college/v82b1imn9naw3y5k89l4kyvfbme8ouaesv.png)
![h'=67.5081\ m](https://img.qammunity.org/2021/formulas/physics/college/24wz01vqw2ez7hvbr7a58o5lvmph7royst.png)
Now the time taken to fall the gross height in course of falling from the top:
![h'=v_y.t'+(1)/(2) g.t'^2](https://img.qammunity.org/2021/formulas/physics/college/ohj6x1zxucldq9jcxka5xu652k68xmkmxa.png)
![67.5081=0+4.9* t'^2](https://img.qammunity.org/2021/formulas/physics/college/gqsv72pxw9oh6autixcvjkl3rya63bdm8w.png)
![t'=3.7118\ s](https://img.qammunity.org/2021/formulas/physics/college/sey2199xxr9dqc0lyw7uze6419ca2sz4bh.png)
Now the total time the projectile spends in the air:
![t_t=t+t'](https://img.qammunity.org/2021/formulas/physics/college/log60095ojdq33byz3ik1juqpm3258jr7u.png)
![t_t=2.576+3.7118](https://img.qammunity.org/2021/formulas/physics/college/g88urtrgthiouuqh82o8wsszvoyionnm7e.png)
![t_t=6.2878\ s](https://img.qammunity.org/2021/formulas/physics/college/xa7e2ea605232fsuzstvnr33zoxcr6f8vz.png)
Now the horizontal component of the initial velocity:
(it remains constant throughout the motion)
![u_x=u.\cos\theta](https://img.qammunity.org/2021/formulas/physics/college/iofa0c4i0zagqffznmjwdhppcr6tqnzadp.png)
![u_x=41* \cos38](https://img.qammunity.org/2021/formulas/physics/college/6nclscumr74w10w273uar43m73aaufptuo.png)
![u_x=32.3084\ m.s^(-1)](https://img.qammunity.org/2021/formulas/physics/college/el0i4ukdrrc3kbhhnvtm3d5j46liyna9hv.png)
Therefore the horizontal distance covered in the total time;
![s=u_x* t_t](https://img.qammunity.org/2021/formulas/physics/college/i000l204mp2xt39kojcb2xruzj23ssu5k0.png)
![s=32.3084* 6.2878](https://img.qammunity.org/2021/formulas/physics/college/i9k8tgld9ord5t5fsev7mgaea6ufwzll2w.png)
![s=203.149\ m](https://img.qammunity.org/2021/formulas/physics/college/t80t3r00pu8spregglwzzibtqd4py4fzhj.png)