Question

With what speed in $\frac{miles}{hr}$ must an object be thrown vertically upwards to reach a height of 91.5 m?
$Takea=9.8\frac{m}{s^2}$

• A. 42.3
• B. 97.4
• C. 94.3
• D. 43.7

Answer 1

The correct option is C

94.3

On reaching maximum height, final velocity = 0

Using third equation of motion, $v^2=u^2+2as$

$0^2=u^2-2\times 9.8\times 91.5$

$u=42.3\frac{m}{s}$

Hence, $u=42.3\times 2.23\frac{miles}{hr}=94.3\frac{miles}{hr}$.