Answer:
Height is different for both .
The time in air are different for both.
Explanation:
Given that initial velocity is same
Lets take initial velocity = u
One at an angle θ.
other at angle 90° − θ.
Also given that their range are same
![R_1=R_2](https://img.qammunity.org/2020/formulas/mathematics/high-school/4jycl9b8kja7ikcd4o967fqdoqi0i28t5y.png)
![R_1=(u^2sin2\theta )/(g)](https://img.qammunity.org/2020/formulas/mathematics/high-school/30uyd9cssqytvsjbhb4n32hpx0x6fey410.png)
![R_2=(u^2sin2(90-\theta))/(g)](https://img.qammunity.org/2020/formulas/mathematics/high-school/ykab6qovkqdl7kfm9im1mx602fyc7ryizt.png)
![R_2=(u^2sin(180-2\theta))/(g)](https://img.qammunity.org/2020/formulas/mathematics/high-school/5obx5whzpvz4v2u1xolmm9yntro3sctjdf.png)
We know that
sin(180° − θ)=sin θ
So
![R_2=(u^2sin2\theta )/(g)](https://img.qammunity.org/2020/formulas/mathematics/high-school/wrawdvl98xas8bufmsvue4y8ie49jlunve.png)
Height in the air
![h_1=(u^2sin^2\theta)/(2g)](https://img.qammunity.org/2020/formulas/mathematics/high-school/kmg9pobdr8jjolobmhln7o5kzy4zrgrwjk.png)
![h_2=(u^2sin^2(90-\theta))/(2g)](https://img.qammunity.org/2020/formulas/mathematics/high-school/fijo463u515fjwp78s9m4lbfkqss14rvjp.png)
We know that
sin(90° − θ)=cos θ
From above we can say that height is different for both .
Time:
![T_1=(2usin\theta)/(g)](https://img.qammunity.org/2020/formulas/mathematics/high-school/xz0aazh1k5wk8l423eocgyiwy7iskpwpv1.png)
![T_2=(2usin(90-\theta))/(g)](https://img.qammunity.org/2020/formulas/mathematics/high-school/u9hfp33afxm3sopyjo1xfgax20jgnu6cms.png)
sin(90° − θ)=cos θ
![T_2=(2ucos\theta)/(g)](https://img.qammunity.org/2020/formulas/mathematics/high-school/7w7h5j2oht45y2hir9v584wtox2pvaikt3.png)
The time in air are different for both.