Question

1.
A projectile is fired horizontally with a velocity of 98 ms -1 from the top of a hill 490m high. Find
(i) the time taken to reach the ground (ii) the distance of the target from the hill and (iii) the angle at which the
projectile hits the ground. (g = 9.8 m/s2)​

Answer 1

Answer:

$t = 10 \\ x = 980 \\ \alpha = 89$

Explanation:

here u=9.8m/s

y=490m

g=9.8m/s²

✒from equation of trejactree

y=1/2(gt²)

490*2=9.8t²

980/9.8=t²

t²=100

t=10sec

✒range is given by

x=ut

x=98*10

x=980m

✒angle made by projectile

here angle Mady by velocity vector with horizontal is

$\beta = \frac{gt}{u}$

since

$\beta = \frac{9.8 \times 10}{98}$

$\beta = 1$

✒since y and u are perpendicular to each other so

$\alpha = 90 - \beta$

$\alpha = 90 - 1$

$\alpha = 89$

✒I hope this is help full for you and please mark it as brilliant