Question

Can anyone solve this question..​

Answer 1

$\leadsto \bf \: given \: \: data \: : \\ \\ \rm \to \: Angle \: of \: projection \: \theta \: = {37}^{ \circ} \\ \\ \rm \to \: initial \: \: velocity \: V_0 = 15 \: m {s}^{ - 1}$

To find :

1. Time for reaching the ball at highest point = t
2. Maximum height = H
3. Horizontal range = R
4. Total time for the ball in air = T

Solution :

${ \boxed{\boxed{\begin{array}{cc} \bf \: (1) \\ \\ \rm \: Time \: for \: reaching \: the \: ball \: at \\ \\ \rm \: highest \: point \: = t =\frac{V_0sin\theta}{g} \\ \\ = \frac{15 \times sin37}{9.8} \\ \\ = 0.921 \: s \end{array}}}}$

${ \boxed{\boxed{\begin{array}{cc} \bf \: (2) \\ \\ \rm \: Maximum \: height = H = \frac{ {V_0}^{2} {sin}^{2} \theta}{2h} \\ \\ = \frac{ {(15)}^{2} \times {(sin37)}^{2} }{2 \times 9.8} \\ \\ = 4.15 \: m \end{array}}}}$

${ \boxed{\boxed{\begin{array}{cc} \bf \: (3) \\ \\ \rm \: Horizontal \: range = R = \frac{ {V_0}^{2}sin2\theta }{g} \\ \\ = \frac{ {(15)}^{2} \times sin(2 \times 37) }{9.8} \\ \\ = 22.06 \: m \end{array}}}}$

${ \boxed{\boxed{\begin{array}{cc} \bf \: (4) \\ \\ \rm \: Total \: time \: in \: air \: = T = \frac{2V_0sin\theta}{g} \\ \\ = \frac{2 \times 15 \times sin37}{9.8} \\ \\ = 1.84 \: s\end{array}}}}$

Answer 2

Answer:

i) Time at which the ball reaches the highest point t = 0.9 s

ii) Maximum height reached = 4.05 m

iii) Horizontal range = 21.6 m

iv) Time for which the ball is in air = 1.8 s

Explanation:

Angle of projection θ = 37°

∴ sin37° = 3/5   and cos37° = 4/5

Initial velocity u = 15 m/s

i) Time of ascent  t = u sinθ/g

                               = 15 x sin37° /10

                               = $\frac{15 X 3}{5 X 10}$

                               = 0.9 s        [ if g is taken as 9.8 then t = 0.91 s ]

Time at which the ball reaches the highest point t = 0.9 s

ii) Max height H$_{max}$ = u² sin²θ / 2g

                                = 15² x (3/5)²/ 2 x 10

                                = (225 x 9)/ (25 x 2 x 10)

                                =81/20

                                = 4. 05 m

Maximum height reached = 4.05 m

iii) Horizontal range R = u²(2 sinθ cosθ)/g

                                    = (225 x 2 x 3/5 x 4/5 )/10

                                     = 9 x 2 x 3 x4 /10

                                     = 21.6 m

iv) Time of flight t$_{f}$ = 2usinθ/g = 2 x time of ascent = 2 x 0.9 = 1.8 s

 ∴ Time for which the ball is in air = 1.8 s